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# 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:**

We know that, in a right triangle, the square of the hypotenuse is equal to the sum of the squares of the other two sides.

In ΔABC, ∠ACB = 90°

D, E are points on AC, BC. Join AE, DE and BD as shown in the diagram above.

In ΔACE,

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

In ΔDCB,

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

Adding equations (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} [Since, in ΔABC, AC^{2} + BC^{2} = AB^{2} and in ΔCDE, CD^{2} + CE^{2} = DE^{2}]

Therefore, AE^{2} + BD^{2} = AB^{2} + DE^{2}

**☛ Check: **NCERT Solutions Class 10 Maths Chapter 6

**Video Solution:**

## 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

NCERT Class 10 Maths Solutions Chapter 6 Exercise 6.5 Question 13

**Summary:**

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

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