Prove that δabc and δedc are similar.

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 postulate.