Question

In triangle ABC,AD is perpendicular to BC and CD=1/3BD.prove 2AB²=2AC²+BC²​

Answer 1

Answer:

2AB²=2AC²+BC²

Step-by-step explanation:

We are given here

AD⊥BC & DB=3CD−−−(1)

so, BC=DB+CD

BC=3CD+CD

so, BC=4CD−−−(2)

Hence it given that,

AB is perpendicular to BC

by the pythagoras theorem

In both triangle △ABD & △ACD

AB^2=AD^2+BD^2

&

AC^2=AD^2+CD^2

∴ AD^2=AB^2−DB^2

&

AD^2=AC^2−CD^2

Compare both AD^2 with each other we get,,

AB^2−DB^2=AC^2−CD^2

AB^2=AC^2+DB^2−CD^2

AB^2=AC^2+(3 CD)*2−CD^2 (from (1))

AB^2=AC^2+9 CD^2−CD^2

AB^2=AC^2+8 CD^2

AB^2=AC^2+8*(4BC )^2 (from (2))

2AB^2=2 AC^2+BC^2 (Let multiply equation with 2)