Question

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

ΔPDC ∼ ΔBEC

Answer 1

In ΔPDC and ΔBEC

∠PDC = ∠BEC = 90°

∠PCD = ∠BCE                   ...(Common angle)

Hence, by using the AA similarity criterion,

ΔPDC ∼ ΔBEC