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

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,

AE² = AC² + CE² (Pythagoras theorem)............... (1)

In ΔDCB,

BD² = CD² + BC² ....................................... (2)

Adding equations (1) and (2)

AE² + BD² = AC² + CE² + CD² + BC²

= AC² + BC² + EC² + CD²

= AB² + DE² [Since, in ΔABC, AC² + BC² = AB² and in ΔCDE, CD² + CE² = DE²]

Therefore, AE² + BD² = AB² + DE²

☛ 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 AE² + BD² = AB² + DE².

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