MAT8 Q2W7D3: Triangle Comparisons & Indirect Proof

By the end of the lesson, you will be able to:

1. Use exterior-angle and side-angle comparisons to order sides and angles in a triangle and justify each comparison in at least 8 items with ≥ 85% accuracy, writing comparisons like $\angle C>\angle B$ and $a>b$.
2. Apply indirect proof (proof by contradiction) to show that a proposed triangle configuration is impossible (e.g., assuming $a\le b$ when given $\angle A>\angle B$) in at least 2 structured proofs with all steps clearly stated.
3. Solve real-world comparison problems (design-navigation-sports) by deciding which opposite side or included angle is larger using triangle-inequality ideas (within-triangle ordering + hinge logic) in at least 5 word problems with complete justifications.

• Exterior angle - an angle formed by extending one side of a triangle.
• Remote interior angles - the two non-adjacent interior angles to a given exterior angle.
• Exterior-angle comparison - in any triangle, an exterior angle is greater than each of its remote interior angles: $\angle \mathrm{ext}>\angle \mathrm{remote1}$ and $\angle \mathrm{ext}>\angle \mathrm{remote2}$.
• Side-angle order (within one triangle) - larger side ↔ larger opposite angle.
• Included angle - the angle formed by two given sides.
• Indirect proof (contradiction) - assume the negation of what you want to prove and derive a contradiction.

A) Identify the included angle
In $\Delta ABC$, for sides $\mathrm{AB}$ and $\mathrm{AC}$, name the included angle.

Show Answer

The angle formed by the two sides at their common endpoint: $\angle A$.

B) Order angles from sides
In $\Delta PQR$, suppose $\mathrm{PQ}>\mathrm{PR}>\mathrm{QR}$. Order the angles from largest to smallest.

Show Answer

Longest side $\mathrm{PQ}$ faces the largest angle $\angle R$; next $\mathrm{PR}$ → $\angle Q$; shortest $\mathrm{QR}$ → $\angle P$. So $\angle R>\angle Q>\angle P$.

C) Hinge logic
Two triangles share $\mathrm{AB}=\mathrm{DE}$ and $\mathrm{AC}=\mathrm{DF}$. If $\angle A>\angle D$, compare $\mathrm{BC}$ and $\mathrm{EF}$.

Show Answer

With the same two adjacent sides, the triangle with the larger included angle has the longer opposite side: $\mathrm{BC}>\mathrm{EF}$.

1) Orientation - why today’s ideas matter

You already know how sides and angles pair up in a triangle: the longest side faces the largest angle, and the shortest side faces the smallest angle. Today you push deeper by using exterior angles and indirect proof to make sharper comparisons without measuring with a protractor or ruler.

Picture a simple tripod. Two legs are fixed to the ground; a third leg is adjustable. When you widen one joint slightly, the angle at that joint grows, so the side across from it changes. This is the heart of within-triangle comparisons. Exterior angles add a new tool: by stepping outside the triangle at one vertex, you gain a comparison that is often quicker than juggling all three interior angles.

Targets for this Explore

• Use the exterior-angle inequality to compare angles quickly.
• Use side-angle order to rank sides or angles in one triangle without calculation.
• Prove comparisons with indirect proof (contradiction) when a direct route is not obvious.
• Apply all of the above to real-life frames, braces, routes, and team spacing.

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2) Exterior angle - the key comparison you can use immediately

Extend one side of a triangle at a vertex. The angle formed outside, adjacent to the interior angle at that vertex, is an exterior angle. In any triangle, an exterior angle is greater than each of its two remote interior angles. More precisely:

• If the exterior angle is at vertex A along side AB extended, then $\angle A\_\mathrm{ext}>\angle B$ and $\angle A\_\mathrm{ext}>\angle C$.

A stronger fact for reasoning is the equality of measures: $$\angle A\_\mathrm{ext}=\angle B+\angle C$$. For inequalities you mostly need that the exterior angle is greater than each remote interior angle.

Quick intuition - when you open the side past the triangle, you add to the interior angle at that vertex, creating a new angle whose measure is the sum of the two interior angles opposite it. A sum of two positive angles must exceed either one alone.

Guided check E1 - In $\Delta PQR$, extend side $\mathrm{PR}$ beyond $R$ to make exterior angle $\angle R\_\mathrm{ext}$. Which angles are the remote interior angles relative to $\angle R\_\mathrm{ext}$, and what are the two basic inequalities you can write immediately?

Show Answer

The remote interior angles are $\angle P$ and $\angle Q$. The inequalities are $\angle R\_\mathrm{ext}>\angle P$ and $\angle R\_\mathrm{ext}>\angle Q$.

Worked walk-through 1 - In $\Delta ABC$, suppose $\angle B=37°$, $\angle C=68°$. Extend side $\mathrm{CA}$ past $A$ to form $\angle A\_\mathrm{ext}$. Since $\angle A\_\mathrm{ext}=\angle B+\angle C$, $\angle A\_\mathrm{ext}=37°+68°=105°$. Therefore $\angle A\_\mathrm{ext}>\angle B$ and $\angle A\_\mathrm{ext}>\angle C$.

Guided check E2 - In $\Delta XYZ$, suppose $\angle Y=51°$, $\angle Z=64°$. Form the exterior angle at $X$ by extending $\mathrm{XZ}$. Compute $\angle X\_\mathrm{ext}$ and record the two inequalities.

Show Answer

$\angle X\_\mathrm{ext}=\angle Y+\angle Z=51°+64°=115°$. Thus $\angle X\_\mathrm{ext}>\angle Y$ and $\angle X\_\mathrm{ext}>\angle Z$.

Mini-summary A - Exterior angles offer instant comparisons: the exterior at a vertex is larger than each remote interior angle. When numbers are given, you can add the two remote interior angles to get the exterior measure.

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3) Side-angle order within one triangle - pairing that never fails

In any triangle, larger side ↔ larger opposite angle. If $a$ is opposite $\angle A$, $b$ opposite $\angle B$, and $c$ opposite $\angle C$, then $$a>b\iff \angle A>\angle B$$ and similar statements cycling through.

Worked walk-through 2 - Given $\Delta MNP$ with sides $\mathrm{MN}=7$, $\mathrm{NP}=12$, $\mathrm{PM}=9$, order the angles: longest side is 12 → largest angle $\angle M$; next is 9 → $\angle N$; smallest is 7 → $\angle P$. So $\angle M>\angle N>\angle P$.

Guided check S1 - Inside $\Delta RST$, suppose $\angle R>\angle T$. Compare sides $\mathrm{ST}$ and $\mathrm{RS}$.

Show Answer

The side opposite the larger angle is longer. Opposite $\angle R$ is $\mathrm{ST}$; opposite $\angle T$ is $\mathrm{RS}$. Therefore $\mathrm{ST}>\mathrm{RS}$.

Mini-summary B - Within any triangle, order sides and angles together: the biggest angle faces the biggest side, and the smallest faces the smallest.

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4) Exterior-angle comparisons + side-angle order = fast rankings

Exterior angles let you compare interior angles that are not obviously related. Combine both tools.

Example blend - In $\Delta ABC$, create the exterior at $A$. If you know $\angle B$ and $\angle C$, then $\angle A\_\mathrm{ext}>\angle B$ and $\angle A\_\mathrm{ext}>\angle C$. Inside the triangle, this often helps locate the largest interior angle among $\angle A$, $\angle B$, $\angle C$, because $\angle A$ and $\angle A\_\mathrm{ext}$ are a linear pair.

Guided check E3 - In $\Delta JKL$, you extend side $\mathrm{JL}$ to form $\angle L\_\mathrm{ext}$. If $\angle J=40°$, $\angle K=61°$, compute $\angle L\_\mathrm{ext}$ and decide which of $\angle J$ or $\angle K$ is larger. Then state which side is longest.

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$\angle L\_\mathrm{ext}=\angle J+\angle K=101°$. Here $\angle K$ is larger than $\angle J$, so the longest side is opposite $\angle K$, namely $\mathrm{JL}$.

Mini-summary C - Exterior angles can help you compare interiors quickly; then use side-angle pairing to locate the longest or shortest side.

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5) Indirect proof - when a direct path is awkward

Sometimes a comparison is not obvious. Indirect proof allows you to assume the opposite of what you want, and then show this leads to something impossible in a triangle, like side-angle order being violated.

Template - Goal: prove $\angle A>\angle B$. Assume the opposite $\angle A\le \angle B$. Then by side-angle pairing, the side opposite $\angle A$ is not longer than the side opposite $\angle B$: $a\le b$. If independent facts already require $a>b$, you have a contradiction. Conclude $\angle A>\angle B$.

Worked walk-through 3 - In $\Delta XYZ$, you know from construction that $\mathrm{XZ}>\mathrm{XY}$. Prove $\angle Y>\angle Z$ indirectly. Assume the opposite $\angle Y\le \angle Z$. Then by side-angle pairing, $\mathrm{XZ}\le \mathrm{XY}$ - contradiction. Hence $\angle Y>\angle Z$.

Guided check IP1 - You want to prove $\mathrm{AB}>\mathrm{AC}$ in $\Delta ABC$. What inequality about angles would you try to prove, and what opposite assumption would you start with?

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Because sides and opposite angles pair, proving $\angle C>\angle B$ will give $\mathrm{AB}>\mathrm{AC}$. Start indirectly by assuming $\angle C\le \angle B$ and derive a contradiction with given information.

Mini-summary D - Indirect proof flips a comparison, then lets the triangle’s rules force a contradiction.

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6) Real-world modeling - quick decisions without exact lengths

Scenario R1 - Folding sign frame
Two struts meet at a hinge at the top of a portable sign. The lower base is the side opposite the hinge angle. If the struts are kept at the same lengths but the hinge opens from $48°$ to $82°$, what happens to the base’s length?

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With adjacent struts fixed, a larger included angle makes the opposite side longer. The base increases as the hinge opens.

Scenario R2 - Two route options
Two hikers walk legs of $1.2$ km and $1.8$ km. Route A turns $35°$ between legs; Route B turns $95°$. Which route has a longer start-to-finish straight line?

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Same two legs; larger included turn gives a longer opposite side. Route B has the longer straight-line distance.

Scenario R3 - Team spacing triangle
Three players form a triangle. Two distances from the pivot are fixed during a drill, while the pivot angle is adjusted. To widen the passing lane opposite the pivot, should the pivot open or close the angle?

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Open the angle. Increasing the included angle increases the opposite side.

Mini-summary E - Design, navigation, and spacing problems often reduce to deciding which angle or side is larger.

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7) Exterior-angle toolkit - four fast uses

1. Compare interiors quickly via an exterior.
2. Spot the largest interior when two are known.
3. Frame impossibility with contradiction when an interior is claimed to exceed its exterior at the same vertex.
4. Chain inequalities: use exterior-inequality plus side-angle pairing to deduce side comparisons.

Guided check E4 - At vertex $A$, a student says the interior $\angle A$ is greater than the exterior $\angle A\_\mathrm{ext}$. Is that ever possible?

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No. $\angle A\_\mathrm{ext}=\angle B+\angle C$, which exceeds either $\angle B$ or $\angle C$ alone. The interior at A cannot exceed its exterior.

Mini-summary F - Exterior-angle facts are one-step inequality engines. Combine them with side-angle order for immediate side rankings.

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8) Worked examples - fully shown solutions

Example W1 - Order angles from sides - In $\Delta QRS$, sides $\mathrm{QR}=8$, $\mathrm{RS}=11$, $\mathrm{SQ}=9$. Conclusion: $\angle Q>\angle R>\angle S$.

Example W2 - Use an exterior angle - In $\Delta ABC$, $\angle B=44°$, $\angle C=59°$. Exterior at A is $103°$, so $\angle A\_\mathrm{ext}>\angle B$, $\angle A\_\mathrm{ext}>\angle C$.

Example W3 - Indirect proof - In $\Delta MNO$, given $\mathrm{MO}>\mathrm{NO}$, assume $\angle N\le \angle M$ leading to $\mathrm{MO}\le \mathrm{NO}$ - contradiction. Hence $\angle N>\angle M$.

Example W4 - Exterior data determines rankings - In $\Delta XYZ$, $\angle X\_\mathrm{ext}=128°$, $\angle Y=53°$. Then $\angle Z=75°$, $\angle X=52°$, so $\mathrm{XY}>\mathrm{XZ}>\mathrm{YZ}$.

Example W5 - Design scenario conversion - Exterior at P is $120°$, $\angle Q=45°$. Then $\angle R=75°$, $\angle P=60°$, and $\mathrm{PQ}>\mathrm{QR}>\mathrm{PR}$.

Example W6 - Indirect with a real constraint - Given $\mathrm{ST}>\mathrm{SU}$, conclude $\angle U>\angle T$ by contradiction.

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9) Practice-like checkpoints inside Explore

Checkpoint C1 - In $\Delta UVW$ with $\mathrm{UV}=10$, $\mathrm{VW}=6$, $\mathrm{WU}=8$, order the angles from largest to smallest.

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Longest side $\mathrm{UV}$ → largest $\angle W$; next $\mathrm{WU}$ → $\angle V$; smallest $\mathrm{VW}$ → $\angle U$. So $\angle W>\angle V>\angle U$.

Checkpoint C2 - In $\Delta ABC$ with $\angle B=50°$ and $\angle C=63°$, form the exterior at $A$ and record the inequalities involving it.

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$\angle A\_\mathrm{ext}=113°$, and $\angle A\_\mathrm{ext}>\angle B$, $\angle A\_\mathrm{ext}>\angle C$.

Checkpoint C3 - You are told $\mathrm{AC}>\mathrm{BC}$. Which angle is larger, and what contradictory assumption would you start with to prove it?

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Larger angle is opposite the larger side: $\angle B>\angle A$. Start by assuming $\angle B\le \angle A$, then derive $\mathrm{AC}\le \mathrm{BC}$, contradicting the given.

Mini-summary H - Practice the full cycle: read, reason, compare, and if needed assume the opposite to reach a contradiction.

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10) Multi-step situations

Situation S1 - Frame with a fixed foot - A folding frame has sides $\mathrm{AB}$ and $\mathrm{AC}$ fixed and meets at hinge $A$. The base is $\mathrm{BC}$. The builder measures an angle outside at $A$ - $\angle A\_\mathrm{ext}$. Can they still decide whether $\mathrm{BC}$ increased or decreased compared to a previous setting?

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Yes. As $\angle A\_\mathrm{ext}$ increases, the interior $\angle A$ decreases because they are a linear pair. The base depends on the interior included angle. A larger exterior means a smaller interior, making $\mathrm{BC}$ shorter.

Situation S2 - Two designs, same legs - Two sign designs use the same legs, but one has a larger exterior angle at the hinge than the other. Which has the longer base?

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Larger exterior at the hinge means smaller interior at the hinge. The design with the smaller interior included angle has the shorter base. So the one with the larger exterior has the shorter base.

Mini-summary I - Exterior and interior at the same vertex move in opposite directions.

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11) Layered reasoning tasks

Layered Task L1 - In $\Delta PQR$, with $\angle P=43°$, $\angle Q=54°$. a) Compute the exterior at $R$. b) Which side is longest?

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a) $\angle R\_\mathrm{ext}=97°$. b) Largest interior is $\angle Q$, so longest side $\mathrm{PR}$.

Layered Task L2 - You are told $\mathrm{AB}>\mathrm{AC}$. Prove $\angle B>\angle A$ by contradiction.

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Assume $\angle B\le \angle A$. Then $\mathrm{AC}\ge \mathrm{AB}$ by side-angle pairing, contradicting $\mathrm{AB}>\mathrm{AC}$. Thus $\angle B>\angle A$.

Mini-summary J - When stuck, flip the target inequality and let side-angle pairing expose the contradiction.

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12) Common pitfalls

• Mixing exterior and interior at the same vertex - they are a linear pair.
• Comparing non-opposite pairs - the size relation always matches the opposite pairing.
• Forgetting strictness - equal angles give equal opposite sides.
• Assuming numbers are required - comparisons are often provable without exact measures.

Guided check P1 - A student says: “Because the exterior at $A$ is bigger than $\angle B$, the side $\mathrm{AC}$ is longer than $\mathrm{BC}$.” Is that correct?

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Not directly. Exterior-angle inequality compares angles, not sides. To compare sides, identify the largest interior angle, then the side opposite that interior is the longest.

Mini-summary K - Use exterior-angle comparisons to rank angles, then switch to side-angle pairing to rank sides.

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13) Extended applications

Construction CA1 - Two brace lengths are fixed in a scissor lift. Choose among interior angles $30°$, $55°$, $95°$ to get the longest opposite side.

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The largest interior included angle yields the longest opposite side. Choose $95°$.

Sports SP1 - Two players keep fixed distances from a pivot. To enlarge the passing lane opposite the pivot, what adjustment increases the lane length?

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Increase the interior angle at the pivot to enlarge the opposite side.

Navigation NV1 - Walk 700 m, turn by $\theta$, walk 900 m. How should $\theta$ change to maximize start-to-end straight line?

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Increase the included turn; larger interior angle yields a longer opposite side.

Mini-summary L - Many field decisions become easy once you know whether to open or close the included angle.

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14) Synthesis set

Synthesis S1 - In $\Delta DEF$, $\angle D=39°$, $\angle E=63°$. a) Compute $\angle F\_\mathrm{ext}$. b) Which side is longest?

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a) $\angle F\_\mathrm{ext}=102°$. b) Largest interior is $\angle E$, so longest side $\mathrm{DF}$.

Synthesis S2 - Indirect - Given $\mathrm{DF}>\mathrm{EF}$, prove $\angle E>\angle D$.

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Assume $\angle E\le \angle D$. Then $\mathrm{DF}\le \mathrm{EF}$, contradicting $\mathrm{DF}>\mathrm{EF}$. Therefore $\angle E>\angle D$.

Mini-summary M - Exterior gives angle comparisons; side-angle pairing converts angle results into side results. Indirect proof closes the loop.

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15) Self-check capsule

1. If $\angle X>\angle Y$, which is longer: $\mathrm{YZ}$ or $\mathrm{XZ}$?
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   $\mathrm{YZ}>\mathrm{XZ}$ because it is opposite the larger angle $\angle X$.
2. If the exterior at $A$ equals $120°$ and one remote interior is $45°$, what is the other remote interior?
   Show Answer

   $75°$ since $45°+x=120°$.
3. If someone claims an interior at $A$ is larger than its exterior at $A$, what is the flaw?
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   The exterior equals the sum of the two remote interiors, so it must exceed either one alone.

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16) Capstone problems

Capstone CP1 - In $\Delta ABC$, $\angle B=41°$, $\angle C=72°$. a) Find $\angle A\_\mathrm{ext}$. b) Which side is the shortest?

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a) $\angle A\_\mathrm{ext}=113°$. b) Smallest interior is $\angle B$, so the shortest side is $\mathrm{AC}$.

Capstone CP2 - Indirect - Given $\mathrm{AB}>\mathrm{CB}$, prove $\angle C>\angle A$.

Show Answer

Assume $\angle C\le \angle A$. Then $\mathrm{AB}\le \mathrm{CB}$, contradicting $\mathrm{AB}>\mathrm{CB}$. Therefore $\angle C>\angle A$.

Capstone CP3 - Exterior + pairing - In $\Delta HIJ$, exterior at $I$ is $128°$ and one remote interior is $50°$. a) Find the other remote interior. b) Which side is longest?

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a) $x+50=128$ → $x=78°$. b) The largest interior is $78°$, so the longest side is opposite that angle.

Mini-summary O - The capstones synthesize everything: an exterior addition, side-angle pairing, and indirect proof.

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References

• Big Ideas Math - Inequalities in one triangle; exterior-angle comparisons.
• MathBitsNotebook - Triangle inequalities, indirect proof outlines.
• CK-12 - Exterior angle theorem and remote interior angles.
• Math is Fun - Angle-side relationships in triangles.
• Open textbooks on Euclidean geometry - Proofs by contradiction in triangle comparisons.

Worked Example 1 - Exterior angle → compare interior angles and sides

In $\Delta ABC$, given $\angle B=42°$, $\angle C=71°$.

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Exterior at A: $\angle A\_\mathrm{ext}=113°$. Interior at A: $\angle A=67°$. Order: $\angle C>\angle A>\angle B$. Sides: $c>a>b$.

Worked Example 2 - Order angles from side lengths

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With sides 7, 12, 10 opposite J, K, L respectively, the order is $\angle J>\angle K>\angle L$.

Worked Example 3 - Indirect proof

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Given $\mathrm{DE}>\mathrm{DF}$, assume $\angle F\le \angle E$ to get $\mathrm{DE}\le \mathrm{DF}$ - contradiction. Hence $\angle F>\angle E$.

Worked Example 4 - Exterior data determines all angle rankings

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With $\angle X\_\mathrm{ext}=128°$ and $\angle Y=53°$, get $\angle Z=75°$, $\angle X=52°$, and side order $\mathrm{XY}>\mathrm{XZ}>\mathrm{YZ}$.

Worked Example 5 - Exterior-to-interior conversion in a design scenario

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Exterior at P $120°$ with $\angle Q=45°$ gives $\angle R=75°$, $\angle P=60°$, and $\mathrm{PQ}>\mathrm{QR}>\mathrm{PR}$.

Worked Example 6 - Indirect proof with a real constraint

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Given $\mathrm{ST}>\mathrm{SU}$, assume $\angle U\le \angle T$ which implies $\mathrm{ST}\le \mathrm{SU}$ - contradiction. Therefore $\angle U>\angle T$.

1. In $\Delta ABC$, $\angle A=46°$, $\angle B=71°$, $\angle C=63°$. Order $b$, $c$, $a$ from longest to shortest.
   Show Answer

   $b>c>a$ by angle order $71°>63°>46°$.
2. In $\Delta JKL$, sides 9, 14, 11. Order $\angle J$, $\angle K$, $\angle L$.
   Show Answer

   $\angle J>\angle K>\angle L$ since 14 > 11 > 9.
3. In $\Delta PQR$, exterior at P is $118°$, $\angle Q=47°$. Find $\angle R$ and interior at P.
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   $\angle R=71°$, $\angle P=62°$.
4. In $\Delta UVW$, if $\angle U>\angle V$, compare $\mathrm{VW}$ and $\mathrm{UW}$.
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   $\mathrm{VW}>\mathrm{UW}$ since opposite the larger angle.
5. In $\Delta XYZ$, exterior at Z is $130°$, $\angle X=58°$. Find $\angle Y$, $\angle Z$, and the longest side.
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   $\angle Y=72°$, $\angle Z=50°$, longest side $\mathrm{XZ}$.
6. Given $\mathrm{AB}>\mathrm{CB}$, prove $\angle C>\angle$

A (outline).

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Assume $\angle C\le \angle A$ → $\mathrm{AB}\le \mathrm{CB}$ contradiction → conclude $\angle C>\angle A$.

In $\Delta MNP$, $\angle M=\angle N$, and $\angle P$ is largest. State relation among $\mathrm{NP}$, $\mathrm{MP}$, $\mathrm{MN}$.

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$\mathrm{MN}>\mathrm{NP}=\mathrm{MP}$.

At vertex $A$, exterior is $124°$. a) Find interior at $A$. b) Can any interior exceed $124°$?

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a) $\angle A=56°$. b) No, the exterior equals the sum of the remote interiors.

Claim check - “Since exterior at $S$ is bigger than $\angle R$, $\mathrm{ST}$ is longer than $\mathrm{RT}$.” Valid?

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Invalid. Compare interiors first, then use side-angle pairing.

In $\Delta CDE$, exterior at D is $110°$ and $\angle C=52°$. a) Find $\angle E$ and interior at D. b) Order $\mathrm{CE}$, $\mathrm{CD}$, $\mathrm{DE}$.

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a) $\angle E=58°$, $\angle D=70°$. b) $\mathrm{CE}>\mathrm{CD}>\mathrm{DE}$.

1. Order sides from angles: $\angle A=48°$, $\angle B=65°$, $\angle C=67°$.
   Show Answer

   $c>b>a$.
2. Order angles from sides: 9, 13, 11.
   Show Answer

   $\angle P>\angle Q>\angle R$.
3. Exterior at M with $\angle N=41°$, $\angle O=58°$.
   Show Answer

   $\angle M\_\mathrm{ext}=99°$, with inequalities to each remote interior.
4. True or False: an interior can exceed its adjacent exterior.
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   False.
5. MCQ - If $\mathrm{UV}=\mathrm{UW}$, which is correct?
   Show Answer

   $\angle V=\angle W$.
6. Indirect proof outline from $\mathrm{AB}>\mathrm{CB}$.
   Show Answer

   Assume $\angle C\le \angle A$ then reach contradiction.
7. Exterior 136° at K → interior?
   Show Answer

   $\angle K=44°$. None can exceed 136°.
8. Two plans 80 m and 120 m: 30°, 70° - which gives longer opposite side?
   Show Answer

   70° gives longer opposite side.
9. Is interior at a vertex ever equal to its exterior of 100°?
   Show Answer

   No, interior is 80°.
10. Which side is longest if angles are 59°, 61°, 60°?
    Show Answer

    Opposite 61°.
11. Exterior 123° at C, remote interior 49°. Find the other remote interior and interior at C.
    Show Answer

    Other remote 74°, interior 57°.
12. Angle order M > N > L. Order sides LN, ML, MN.
    Show Answer

    $\mathrm{LN}>\mathrm{ML}>\mathrm{MN}$.
13. As exterior at A increases with equal sides AD = AE, what happens to DE?
    Show Answer

    Interior at A decreases, so DE shortens.
14. Exterior 112° at J, $\angle H=47°$. Find $\angle I$ and the longest side.
    Show Answer

    $\angle I=65°$, longest side $\mathrm{HJ}$.
15. Short proof: if $\angle S>\angle T$, compare TV and SV.
    Show Answer

    Opposite the larger angle is longer: TV > SV.

Extension 1 - Design Optimization with a Hinge

Isosceles stand with equal legs AB and AC, base BC. Compare base lengths at 30°, 55°, 80° without computing exact lengths.

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Larger included angle → longer opposite side. Short → long: 30° < 55° < 80°.

Extension 2 - Indirect Proof: Spotting an Impossible Claim

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If PQ = PR but $\angle Q>\angle R$, then PR > PQ by side-angle pairing, contradicting PQ = PR.

Extension 3 - Exterior Angle Chain Reasoning

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With exterior 126° at X and $\angle Y=46°$, get $\angle Z=80°$, $\angle X=54°$, and sides $\mathrm{XY}>\mathrm{YZ}>\mathrm{ZX}$.

Extension 4 - Route Planning Threshold

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With sides 3 km and 4 km, the smallest angle that guarantees at least 5 km is 90°, since right triangle gives exactly 5; larger angles give more.

Extension 5 - Create & Solve Your Own Indirect Proof

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Sample: Given $\mathrm{AB}>\mathrm{CB}$, assume $\angle C\le \angle A$ → $\mathrm{AB}\le \mathrm{CB}$ contradiction → conclude $\angle C>\angle A$.

Write in your notebook. Use the 3-2-1 format to reflect on today’s learning (exterior angles, side-angle order, and indirect proof). Be specific and refer to at least one problem you solved.

1. 3 insights - three things you now understand about comparing sides and angles in one triangle.
2. 2 connections - two ways today’s ideas connect to real situations or earlier lessons.
3. 1 question - one question you still have or one step in an indirect proof you want to practice again.

(Optional) Sketch a small triangle and label: one interior angle, the adjacent exterior, and the two remote interiors. Add a one-sentence note about how their sizes compare.