Prove that ΔABC and ΔEDC are similar.

Question

ABC and DEC where angles A and E are right angles, AC equals 4, AB equals 3, BC equals 5, DC equals 15, DE equals 9, and CE equals 12

15 over 5 equals 12 over 4 equals 9 over 3 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SSS Similarity Postulate.
∠DCE is congruent to ∠CBA by the Vertical Angles Theorem and 15 over 5 equals 12 over 4 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SSS Similarity Postulate.
∠E and ∠B are right angles and, therefore, congruent since all right angles are congruent. 9 over 4 and 12 over 3 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SAS Similarity Postulate.
15 over 4 equals 12 over 5 equals 9 over 3 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SAS Similarity Po

Answer 1

To prove that triangles $\mathrm{\Delta }ABC$ and $\mathrm{\Delta }EDC$ are similar, we can utilize the concept of side lengths and angles.

Given Information:

• $\mathrm{\angle }A$ and $\mathrm{\angle }E$ are both right angles (each measuring ${90}^{\circ }$).
• Side lengths:
  • $AC=4$
  • $AB=3$
  • $BC=5$
  • $DC=15$
  • $DE=9$
  • $CE=12$

Step 1: Verify Corresponding Sides Are Proportional

We need to establish the ratios of the corresponding sides of the triangles:

1. Ratio of $AC$ and $CE$: $$\frac{CE}{AC}=\frac{12}{4}=3$$

2. Ratio of $AB$ and $DE$: $$\frac{DE}{AB}=\frac{9}{3}=3$$

3. Ratio of $BC$ and $DC$: $$\frac{DC}{BC}=\frac{15}{5}=3$$

Conclusion from Step 1

Since all the ratios of corresponding sides are equal: $$\frac{AC}{CE}=\frac{AB}{DE}=\frac{BC}{DC}=3$$ This shows that the corresponding sides are proportional.

Step 2: Establish Angle Congruence

1. We know $\mathrm{\angle }A\cong \mathrm{\angle }E$ (both are ${90}^{\circ }$).
2. By the Vertical Angles Theorem, $\mathrm{\angle }DCE\cong \mathrm{\angle }CBA$ since they are vertical angles.

Step 3: Triangle Similarity Conclusion

Now we can conclude using the Angle-Angle (AA) Similarity Postulate because:

• $\mathrm{\angle }A\cong \mathrm{\angle }E$ (right angles)
• $\mathrm{\angle }DCE\cong \mathrm{\angle }CBA$ (vertical angles)

Since two angles of triangle $\mathrm{\Delta }ABC$ are congruent to two angles of triangle $\mathrm{\Delta }EDC$, the triangles are similar.

Final Result

Thus, we can say: $$\mathrm{\Delta }ABC\sim \mathrm{\Delta }EDC$$

This similarity comes from both the proportionality of the sides and the congruence of the angles.