Question

the perpendicular from A on side BC of a triangle ABC intersect BC at D such that DB =3CD proove that 2AB2=2Ac2+BC2​

Answer 1

Step-by-step explanation:

SOLUTION:

BC=BD+CD...(B-C-D)

BC=3CD+CD... (Given)

BC=4CD

BC^2=16CD^2-----*1

Now,

In triangle ADC;

AC^2=AD^2+CD^2

AD^2=AC^2-CD^2-----*2

Similarly,

In triangle ADB;

AD^2=AB^2-BD^2-----*3

From *2 & *3;

AC^2-CD^2=AB^2-BD^2

AC^2-CD^2=AB^2-(3CD)^2

AC^2=AB^2-9CD^2+CD^2

AC^2=AB^2-8CD^2

2AC^2=2AB^2-16CD^2

2AC^2=2AB^2-BC^2..... (From *1)

Answer 2

Answer:

See the attachment. Hope it helps uh!