# 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:**

In a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides. [Pythagoras theorem]

In ΔABC, ∠ACB = 90°

D, E are points on AC, BC .

Join AE, DE and BD

In ΔACE

AE^{2} = AC^{2} + CE^{2} (Pythagoras theorem)............... (1)

In ΔDCB

BD^{2} = CD^{2} + BC^{2} ....................................... (2)

Adding (1) and (2)

AE^{2} + BD^{2} = AC^{2} + CE^{2} +CD^{2} + BC^{2}

= AC^{2} + BC^{2} + EC^{2} + CD^{2}

= AB^{2} + DE^{2}

[In ΔABC, ∠C = 90° ⇒ AC^{2} + BC^{2} = AB^{2} and

In ΔCDE, ∠DCE = 90° ⇒ CD^{2} + CE^{2} = DE^{2}

⇒ AE^{2} + BD^{2} = AB^{2} + DE^{2}

**Video Solution:**

## 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}

### NCERT Class 10 Maths Solutions - Chapter 6 Exercise 6.5 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}

In the above D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C. Hence proved that AE^{2} + BD^{2} = AB^{2} + DE^{2}