Question

Prove that in a right angled triangle square of the hypotenus is equal to sum of the square of other two side

Answer 1

Figure is in the attachment


Given:

A right angled ∆ABC, right angled at B



To Prove- AC²=AB²+BC²



Construction: draw perpendicular BD onto the side AC .


Proof:

We know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, than triangles on both sides of the perpendicular are similar to the whole triangle and to each other.


We have

△ADB∼△ABC. (by AA similarity)


Therefore, AD/ AB=AB/AC


(In similar Triangles corresponding sides are proportional)



AB²=AD×AC……..(1)


Also, △BDC∼△ABC


Therefore, CD/BC=BC/AC


(in similar Triangles corresponding sides are proportional)


Or, BC²=CD×AC……..(2)



Adding the equations (1) and (2) we get,



AB²+BC²=AD×AC+CD×AC


AB²+BC²=AC(AD+CD)


( From the figure AD + CD = AC)


AB²+BC²=AC . AC



Therefore, AC²=AB²+BC²

This theroem is known as Pythagoras theroem...

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Hope this will help you.....

Answer 2

AC²=AB²+BC²

Given: A right angled ∆ABC, right angled at B

To Prove: AC²=AB²+BC²

Construction: Draw perpendicular BD onto the side AC .

Proof:

We know that if a perpendicular is drawn from the vertex of a right angle of a right angled triangle to the hypotenuse, than triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

We have

△ADB∼△ABC. (by AA similarity)

Therefore, AD/ AB=AB/AC

(In similar Triangles corresponding sides are proportional)

AB²=AD×AC……..(1)

Also, △BDC∼△ABC

Therefore, CD/BC=BC/AC

(in similar Triangles corresponding sides are proportional)

Or, BC²=CD×AC……..(2)

Adding the equations (1) and (2) we get,

AB²+BC²=AD×AC+CD×AC

AB²+BC²=AC(AD+CD)

( From the figure AD + CD = AC)

AB²+BC²=AC . AC

Therefore, AC²=AB²+BC²

Hope this helps you!!!