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Triangles

Question 7:

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

(i) ΔAEP ∼ ΔCDP

(ii) ΔABD ∼ ΔCBE

(iii) ΔAEP ∼ ΔADB

(v) ΔPDC ∼ ΔBEC

Answer:

(i)

In ΔAEP and ΔCDP,

∠AEP = ∠CDP (Each 90°)

∠APE = ∠CPD (Vertically opposite angles)

Hence, by using AA similarity criterion,

ΔAEP ∼ ΔCDP

(ii)

In ΔABD and ΔCBE,

∠ADB = ∠CEB (Each 90°)

∠ABD = ∠CBE ...

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