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.