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

GPT-4o mini · Answer

To prove that triangles $ \Delta ABC $ and $ \Delta EDC $ are similar, we can utilize the concept of side lengths and angles. ### Given Information: - $ \angle A $ and $ \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 $ \angle A \cong \angle E $ (both are $ 90^\circ $). 2. By the Vertical Angles Theorem, $ \angle DCE \cong \angle CBA $ since they are vertical angles. ### Step 3: Triangle Similarity Conclusion Now we can conclude using the **Angle-Angle (AA) Similarity Postulate** because: - $ \angle A \cong \angle E $ (right angles) - $ \angle DCE \cong \angle CBA $ (vertical angles) Since two angles of triangle $ \Delta ABC $ are congruent to two angles of triangle $ \Delta EDC $, the triangles are similar. ### Final Result Thus, we can say: $$ \Delta ABC \sim \Delta EDC $$ This similarity comes from both the proportionality of the sides and the congruence of the angles.