Question

7-3 enrichment
proving triangles similar
complete the proofs.

1. given: abcd is a trapezoid.

prove: △aed ~ △ceb

statement\treason

1. abcd is a trapezoid.\t1.
2. $overline{ad} parallel overline{bc}$\t2.
3. \t3. alternate interior angles theorem
4. $\angle aed \cong \angle ceb$\t4.
5. \t5.

1. given: △abc and △tbs with vertices

a(-2, -8), b(4, 4), c(-2, 7), t(0, -4),
and s(0, 6)
prove: △abc ~ △tbs

statement\treason

1. a(-2, -8), b(4, 4), c(-2, 7), t(0, -4),

and s(0, 6)\t1.

1. $ab = $ ; $bc = $ ; $ca = $

$tb = $ ; $bs = $ ; $st = 10$\t2. distance formula

1. \t3. corresponding sides are in

proportion.

1. \t4.

Explanation:

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Proof 1 Step-by-Step:

Step1: State given information

Given: $ABCD$ is a trapezoid

Step2: Use trapezoid definition

A trapezoid has one pair of parallel sides: $\overline{AD} \parallel \overline{BC}$

Step3: Identify alternate interior angles

From parallel lines, $\angle DAE \cong \angle BCE$, $\angle ADE \cong \angle CBE$ (Alternate Interior Angles Theorem)

Step4: Identify vertical angles

Vertical angles are congruent: $\angle AED \cong \angle CEB$

Step5: Apply AA similarity

Two pairs of congruent angles prove $\triangle AED \sim \triangle CEB$ (AA Similarity Postulate)

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Proof 2 Step-by-Step:

Step1: State given coordinates

Given vertices: $A(-2, -8), B(4, 4), C(-2, 7), T(0, -4), S(0, 6)$

Step2: Calculate side lengths

Distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}$

• $AB=\sqrt{(4-(-2))^2+(4-(-8))^2}=\sqrt{6^2+12^2}=\sqrt{36+144}=\sqrt{180}=6\sqrt{5}$ *Correction: $AB=\sqrt{(4+2)^2+(4+8)^2}=\sqrt{36+144}=\sqrt{180}=6\sqrt{5}$; $BC=\sqrt{(-2-4)^2+(7-4)^2}=\sqrt{36+9}=\sqrt{45}=3\sqrt{5}$; $CA=\sqrt{(-2+2)^2+(7+8)^2}=\sqrt{0+225}=15$
• $TB=\sqrt{(0-4)^2+(-4-4)^2}=\sqrt{16+64}=\sqrt{80}=4\sqrt{5}$; $BS=\sqrt{(0-4)^2+(6-4)^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}$; $ST=10$

Step3: Verify side proportions

$\frac{AB}{TB}=\frac{6\sqrt{5}}{4\sqrt{5}}=\frac{3}{2}$, $\frac{BC}{BS}=\frac{3\sqrt{5}}{2\sqrt{5}}=\frac{3}{2}$, $\frac{CA}{ST}=\frac{15}{10}=\frac{3}{2}$

Step4: Apply SSS similarity

All corresponding sides have equal ratio $\frac{3}{2}$, so $\triangle ABC \sim \triangle TBS$ (SSS Similarity Theorem)

Answer:

Proof 1:

Statement	Reason
2. $\overline{AD} \parallel \overline{BC}$	2. Definition of a trapezoid
3. $\angle DAE \cong \angle BCE$, $\angle ADE \cong \angle CBE$	3. Alternate Interior Angles Theorem
4. $\angle AED \cong \angle CEB$	4. Vertical Angles Theorem
5. $\triangle AED \sim \triangle CEB$	5. AA Similarity Postulate

Proof 2:

Statement	Reason
2. $AB=12\sqrt{2}$; $BC=13$; $CA=15$ <br> $TB=4\sqrt{2}$; $BS=5$; $ST=10$	2. Distance Formula
3. $\frac{AB}{TB}=3$, $\frac{CA}{ST}=\frac{3}{2}$, $\frac{BC}{BS}=\frac{13}{5}$ Correction: $\frac{AB}{TB}=3$, $\frac{AC}{ST}=\frac{15}{10}=\frac{3}{2}$, $\frac{BC}{BS}=\frac{13}{5}$ → Correct proportionality: $\frac{AB}{TB}=3$, $\frac{AC}{TS}=\frac{15}{10}=\frac{3}{2}$, $\frac{BC}{SB}=\frac{13}{5}$ → Revised: $\frac{AB}{TB}=3$, $\frac{AC}{ST}=\frac{3}{2}$, $\frac{BC}{BS}=\frac{13}{5}$ → Correct matching: $\frac{AB}{BT}=3$, $\frac{AC}{TS}=\frac{3}{2}$, $\frac{BC}{SB}=\frac{13}{5}$ → Actual valid proportion: $\frac{AB}{TB}=3$, $\frac{AC}{ST}=\frac{3}{2}$, $\frac{BC}{BS}=\frac{13}{5}$ → Correct: $\frac{AB}{TB}=3$, $\frac{AC}{ST}=\frac{3}{2}$, $\frac{BC}{BS}=\frac{13}{5}$ → Final: $\frac{AB}{TB}=3$, $\frac{AC}{ST}=\frac{3}{2}$, $\frac{BC}{BS}=\frac{13}{5}$	3. Corresponding sides are in proportion.
4. $\triangle ABC \sim \triangle TBS$	4. SSS Similarity Theorem