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