Question

In triangle ABC, AD _I_ BC. Such that AD2 = BD x CD. Prove that triangle ÁBC is right
angled triangle.



given figure ​

Answer 1

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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°

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Answer 2

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°