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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