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Question

Question 13
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.

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Solution


Applying Pythagoras theorem in ΔACE, we obtain
AC2+CE2=AE2(1)
Applying Pythagoras theorem in ΔBCD, we obtain
BC2+CD2=BD2(2)
Using equation (1) and equation (2), we obtain
AC2+CE2+BC2+CD2=AE2+BD2(3)
Applying Pythagoras theorem in ΔCDE, we obtain
DE2=CD2+CE2
Applying Pythagoras theorem in ΔABC, we obtain
AB2=AC2+CB2
Putting the values in equation (3), we obtain
DE2+AB2=AE2+BD2

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