Log in to add a comment

## Answers

AD is perpendicular to BC, therefore BD=DC=1/2BC=1/2AB

In ΔABC, AB²=AD²+BD² .....(1)

In ΔACD, AC²=AD²+DC² .......(2)

(1)+(2)

AB²+AC²=AD²+BD²+AD²+DC²

=2AD²+BD²+DC²

AB²+AB²=2AD²+1/4AB²+1/4AB²

2AB²=2AD²+1/2AB²

(2-1/2AB²)=2AD²

3/2AB²=2AD²

3AB²=4AD²

Hope it helps.

AB=BC=AC

LOOK INTO ΔABD

*ANGLES ARE 30,60,90*

*SIDES RATIO=X:X√3:2X*

AB=2X

AB²=4X²

AND AD=X√3

AD²=3X²

3AB²=4X²*3=12X²

SO WE CAN SAY THAT

AD²*4=12X²

(3X²*4)=12X²

WE PROVED THAT 4AD²=3AB²

AB=2X

AB²=4X²

AND AD=X√3

AD²=3X²

3AB²=4X²*3=12X²

SO WE CAN SAY THAT

AD²*4=12X²

(3X²*4)=12X²

WE PROVED THAT 4AD²=3AB²