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

For Competitive Exams

Question 13 D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Prove that $AE^2 + BD^2 = AB^2 + DE^2$.

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

$AC^2 + CE^2 = AE^2$ ………………………………………….(i)

In ΔBCD, by Pythagoras theorem, we get

$BC^2 + CD^2 = BD^2$ ………………………………..(ii)

From equations (i) and (ii), we get,

$AC^2 + CE^2 + BC^2 + CD^2 = AE^2 + BD^2$ …………..(iii)

In ΔCDE, by Pythagoras theorem, we get

$DE^2 = CD^2 + CE^2$

In ΔABC, by Pythagoras theorem, we get

$AB^2 = AC^2 + CB^2$

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

$DE^2 + AB^2 = AE^2 + BD^2$

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