Question

in a right angle triangle right angle at A AD perendicular to BC then BD/DC=....​

Answer 1

$\huge\underline\mathcal\red{ANSWER♡}$

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°