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

Answer:

Applying Pythagoras theorem in ΔACE, we obtain

AC² + CE² = AE²       (1)

Applying Pythagoras theorem in ∆BCD, 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 ∆CDE, we obtain

DE² = CD² + CE²

Applying Pythagoras theorem in ∆ABC, we obtain

AB² = AC² + CB²

Putting the values in equation (3), we obtain

DE² + AB² = AE² + BD²

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