#### Question

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

ΔAEP ∼ ΔADB

#### Solution

In ΔAEP and ΔADB,

∠AEP = ∠ADB (Each 90°)

∠PAE = ∠DAB (Common)

Hence, by using AA similarity criterion,

ΔAEP ∼ ΔADB

Is there an error in this question or solution?

Solution In the following figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that Concept: Criteria for Similarity of Triangles.