Question

$Prove \: that \: in \: right \: angled \: triangle, \: \\ height ² \: +base ² \: = hypotenious ²$
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Answer 1

Answer:

Step-by-step explanation:

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Answer 2

*Answer:

Given:- A right angled triangle ABC, right angle at B

To prove:- AC² = AB ² + BC²

Proof:- Draw a perpendicular BD from B to AC

In △ABC and △ABD

∠ADB=∠ABC=90°

∠DAB=∠BAC ..... (Common angle)

∴△ABC∼△ABD ...... (Using AA similarity criteria)

Now ,

$= > \frac{ad}{ab} = \frac{ab}{ac} \: (corresponding \: sides \: are \: equal)$

⇒AB ² = AD×AC ...... (i)

Similarly, △ABC ∼△BDC

∴ BC² = CD×AC ...... (ii)

Adding equations (i) and (ii), we get

AB² + BC² = AD × AC + CD × AC

⇒AB² + BC² = AC (AD+CD)

⇒AB² + BC² = AC × AC

⇒AB² + BC² = AC² [hence proved]..