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

ΔPDC ∼ ΔBEC

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#### Solution

In ΔPDC and ΔBEC,

∠PDC = ∠BEC (Each 90°)

∠PCD = ∠BCE (Common angle)

Hence, by using AA similarity criterion,

ΔPDC ∼ ΔBEC

Concept: Criteria for Similarity of Triangles

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