Question

In triangle ABC AD is perpenicular to BC and AD^=BD x CD. prove that AB^+ AC ^= (BD+ CD)^ ​

Answer 1

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°

Answer 2

Answer:

See figure of question in attachment

Step-by-step explanation:

In The triangle ABC

AD is perpenicular to BC &

AD^2=BD x DC

The triangle ABC is right angle triangle

By Pythagoras theorem

AB^2 + AC^2 = BC^2

AB^2 + AC^2 = (BD+ CD)^ 2...... (B-D-C)

AB^2 + AC^2 = (BD+ CD)^ 2 IS PROVED