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

ΔABD ∼ ΔCBE

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#### Solution

In ΔABD and ΔCBE,

∠ADB = ∠CEB (Each 90°)

∠ABD = ∠CBE (Common)

Hence, by using AA similarity criterion,

ΔABD ∼ ΔCBE

Concept: Criteria for Similarity of Triangles

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