Question

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 1

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MATHS

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

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ANSWER

Given:- △ABC, right angled at C.

Two points D and E are on the sides AC and AB.

To prove:- AE 2

+BD 2

=AB 2

+DE 2

Construction:- Join the point D with B and E and point E with A.

Proof:-

Using pythagoras theorem,

Hypotenuse 2

=Perpendicular 2

+Base 2

In △ACE,AE 2

=AC 2

+CE 2

.....(1)

In △BCD,BD 2

=DC 2

+BC 2

.....(2)

In △ABC,AB 2

=AC 2

+BC 2

.....(3)

In △DCE,DE 2

=DC 2

+CE 2

.....(4)

Now, A2

+BD 2

=AB 2

+DE 2 AC 2

+CE 2

+DC 2

+BC 2

=AC 2

+BC 2

+DC 2

+CE 2

L.H.S. = R.H.S.

Hence proved.

Answer 2

Given :-

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.

To Find :-

Prove that AE² + BD² = AB² + DE²

Solution :-

Given that,

D and E are points on the sides CA and CB respectively of a triangle ABC right angled at C.

By Pythagoras theorem, ΔACE,

$\sf AC^{2} + CE^{2} = AE^{2} \qquad ...(1)$

By Pythagoras theorem, ΔBCD,

$\sf BC^{2} + CD^{2} = BD^{2} \qquad...(2)$

From equations (1) and (2),

$\sf AC^{2} + CE^{2} + BC^{2} + CD^{2} = AE^{2} + BD^{2} \qquad. ..(3)$

By Pythagoras theorem, ΔCDE,

$\sf DE^{2} = CD^{2} + CE^{2}$

By Pythagoras theorem, ΔABC,

$\sf AB^{2} = AC^{2} + CB^{2}$

Substituting the values in equation (3)

$\sf DE^{2} + AB^{2} = AE^{2} + BD^{2}$