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Question
In the following figure, altitudes AD and CE of ΔABC intersect each other at the point P. Show that:
ΔAEP ∼ ΔCDP
Solution
In ΔAEP and ΔCDP,
∠AEP = ∠CDP ...(Each 90°)
∠APE = ∠CPD ...(Vertically opposite angles)
Hence, by using AA similarity criterion,
ΔAEP ∼ ΔCDP
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