D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that AE2 + BD2 = AB2 + DE2.

Asked by Pragya Singh | 1 year ago |  39

##### Solution :-

Given, D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.

By Pythagoras theorem in ΔACE, we get

AC2 + CE2 = AE2 ………….(i)

In ΔBCD, by Pythagoras theorem, we get

BC2 + CD2 = BD2 ……………..(ii)

From equations (i) and (ii), we get,

AC2 + CE2 + BC2 + CD2 = AE2 + BD2 ………..(iii)

In ΔCDE, by Pythagoras theorem, we get

DE2 = CD2 + CE2

In ΔABC, by Pythagoras theorem, we get

AB2 = AC2 + CB2

Putting the above two values in equation (iii), we get

DE2 + AB2 = AE2 + BD2.

Answered by Abhisek | 1 year ago

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