Question

please help me...here...

Answer 1

To PROVE THAT : AC + AD = BC

Since AB = AC = a , <A = 90°

=> BC² = a² + a² = 2a² ( by Pythagoras law)

=> BC = √2a

Now, since CD is bisector of < C

=> AC/ BC = AD/BD ( by angle bisector theorem)

=> a/√2a = AD/BD

=> 1 /√2 = AD/BD = x/√2x

=> AB = x + √2x = x(1+√2)

So, AC = x(1+√2)

By Pythagoras law

AB² + AC² = BC²

=> x²(1+√2)² + x² (1+√2)² = BC²

=> BC = √{2x²(1+√2)²}

=> BC = √2 x ( 1+√2)

=> BC = √2x + 2x ………….. (1) (RHS)

Now, since AC = x + √2x

& AD = x

=> AC + AD = x+ √2x + x

=> AC + AD = √2x + 2x ………… ( 2) LHS

therefore LHS=RHS

I hope this will help you.