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Question:

In Fig, altitudes AD and CE of ΔABC intersect each other at the point P. Show that: (i) ΔAEP ≅ ΔCDP (ii) ΔABD ≅ ΔCBE (iii) ΔAEP ≅ ΔADB (iv) ΔPDC ≅ ΔBEC

Solution:

In ΔAEP and ΔCDP, ∠APE = ∠CPD (Vertically opposite angle) ∠AEP = ∠CDP = 90⁰
∴ By AA criterion of similarity, ΔAEP ≅ ΔCDP
In ΔABD and ΔCBE ∠ADB = ∠CEB = 90⁰ ∠B is common
∴ By AA criterion of similarity, ΔABD ≅ ΔCBE
In ΔAEP and ΔADB ∠AEP = ∠ADB = 90⁰ ∠A is common
∴ By AA criterion of similarity, ΔAEP ≅ ΔADB
ΔPDC and ΔBEC ∠PDC = ∠BEC = 90⁰ ∠C is common
∴ By AA criterion of similarity, ΔPDC ≅ ΔBEC