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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