Question

In tri.ABC , angleA is acute . BD AND CE are parpendiculars on AC and AB respectively. prove that AB×AE = AC×AD

Answer 1

This question will be solved by B.P.T theorem

Answer 2

Consider the △ABC,

So, we have to prove that, AB×AE=AC×AD

Now, consider the △ADB and △AEC,

∠A=∠A [common angle for both triangles]

∠ADB=∠AEC [both angles are equal to 90⁰]

△ADB∼△AEC

So, AB/AC=AD/AE

By cross multiplication we get,

AB×AE=AC×AD.