Question

In the following figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:

ΔAEP ∼ ΔADB

Answer 1

In ΔAEP and ΔADB,

∠AEP = ∠ADB       ...(Each 90°)

∠PAE = ∠DAB       ...(Common)

Hence, by using AA similarity criterion,

ΔAEP ∼ ΔADB