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

Given: ΔACB is a right angles triangle at C.
Construction: Join AE, BD and ED.
To Prove: AE² + BD² = AB² + DE²
Proof:
In ΔACE
By pythagoras theorem, AE² = EC² + AC² (1)
In ΔBCD
By pythagoras theorem, BD² = BC² + CD² (2)
In ΔECD
By pythagoras theorem, ED² = EC² + DC² (3)
In ΔABC
By pythagoras theorem, AB² = BC² + AC² (4)
Adding 1) and 2) we get,
AE² + BD² = EC² + AC² + BC² + CD² (5)
From 3), 4) and 5) we get,
AE² + BD² = AB² + DE² [hence proved]