Question

Prove that in a right triangle, the square on the hypotenuse is

equal to the sum of the squares on the other two sides.

In ∆ABC,

AD  BC

and CD=

3

1

BD. Using

the above theorem, Prove that

2 2 2

2AB  2AC  BC​

Answer 1

Given;

A right angles triangle ABC

To find:

Prove that in a right angle triangle, the square of the hypotenuse is equal to the sum of squares of the other two sides.

Solution:

Given: ABC is a triangle in which ∠ABC=90°

Construction: Draw BD⊥AC.

Proof:

In △ADB and △ABC

∠A=∠A              [Common angle]

∠ADB=∠ABC      [Each 90°]

△ADB∼△ABC    [A−A Criteria]

So,  

AB /AD  =  AC /AB

Now, AB ^2  =AD×AC             ..........(1)

Similarly,

BC ^2  =CD×AC               ..........(2)

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

AB^2  +BC ^2  =AD×AC+CD×AC

=AC(AD+CD)

=AC×AC

∴AB^2 +BC ^2  =AC ^2

Hence proved.