PLEASE HELP ME Prove that ΔABC and ΔEDC are similar.

triangles 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 4 equals 12 over 5 equals 9 over 3 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SAS Similarity Postulate.

∠E and ∠A are right angles; therefore, these angles are congruent since all right angles are congruent. 15 over 5 equals 12 over 4 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SSS Similarity Postulate.

∠E and ∠A are right angles; therefore, these angles are congruent since all right angles are congruent. 12 over 4 equals 9 over 3 shows the corresponding sides are proportional; therefore, ΔABC ~ ΔEDC by the SAS Similarity Postulate.

∠DCE is congruent to ∠BCA 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.