Question

In triangle abc ad is perpendicular to BC prove that a b square + CD square = B d square + AC square

Answer 1

In the image above:

ABD is a right angled traingle
and AB is the hypotaneous

(AB)^2 = (AD)^2 + (BD)^2

(AB)^2 - (BD)^2 = (AD)^2 ---------(i)


now,
ACD is also a right angled traingle
(AC)^2 = (AD)^2 + (DC)^2
(AC)^2 - (DC)^2 = (AD)^2 ----------(ii)

(i) = (ii)

(AB)^2 - (BD)^2 =(AC)^2 - (DC)^2
(AB)^2 +(DC)^2 = (AC)^2 + (BD)^2

Answer 2

Hey!!!

In triangle ABC,
Triangle ABD and Triangle ACD are right angle triangle. So by using Pythagoras theorem,
Triangle ABD,
AB^2 = AD^2 + BD^2 ----------------(1)
Triangle ACD,
AC^2 = AD^2 + CD^2 ---------------(2)
Subtract both equations,
AB^2 - AC^2 = AD^2 + BD^2 - ( AD^2 + CD^2)
AB^2 - AC^2 = AD^2 + BD^2 - AD^2 - CD^2
AB^2 - AC^2 = BD^2 - CD^2
AB^2 - BD^2 = AC^2 - CD^2
Hence proved.

Hope it helps u....