Question

prove that In a right-angled traingle, the square on the hypotenuse is equal to the sum of squares on the other two sides

Answer 1

Let,
In a right angled triangle ABC, angle B=90°
construction:- draw BD perpendicular to AC

proof:-
∆ABC~∆BDC
by B.P.T
BC/AC=DC/BC
BC²=AC×DC -- eq1

also, ∆ABC~∆ADB
by B.P.T
AB/AC=AD/AB
AB²=AC×AD-- eq2

so, add eq 1 and 2 AB²+BC²=(AC×DC)+ (AC×AD)

AB²+BC²=AC(DC+AD)
AB²+BC²=AC×AC

W.K.T, AC=AD+DC

SO, AB²+BC²=AC²

Hence proved

B.P.T= basic proportionality theorem

W.K.T= we know that

~ = is similar to

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!!!