Question

In an isosceles AABC, D is a point on BC such that ADI BC, then
(A) AB? - AD2 = BD.DC
(B) AB2- AD2 = BD2 - DC2
(C) AB? + AD2 = BD.DC
(D) AB? + AD? = BD2 - DC2​

Answer 1

Step-by-step explanation:

Given : In triangle ABC , AD is perpendicular to BC and AD² = BD.DC

To prove : BAC = 90°

Proof : in right triangles ∆ADB and ∆ADC

So, Pythagoras theorem should be apply ,

Then we have ,

AB² = AD² + BD² ----------(1)

AC²= AD²+ DC² ---------(2)

AB² + AC² = 2AD² + BD²+ DC²

= 2BD . CD + BD² + CD² [ ∵ given AD² = BD.CD ]

= (BD + CD )² = BC²

Thus in triangle ABC we have , AB² + AC²= BC²

hence triangle ABC is a right triangle right angled at A

∠ BAC = 90°