Question

ABC is a right triangle right-angled at B.Let D and E be any points on AB and BC respectively.Prove that AE²+CD²=AC²+DE²
(class 10 CBSE SAMPLE PAPER 2017-18 MATHS)

Answer 1

ΔABE is a right triangle, right angled at B

AB²+BE² = AE²……………..(1)

(by the Pythagoras theorem)


ΔDBC is a right triangle, right angled at B

DB²+BC² = CD²…………….(2)
(by the Pythagoras theorem


Adding eq 1 & 2


AE²+CD²= (AB²+BE²)+(BD²+BC²)


AE²+CD²= (AB²+BC²)+(BE²+BD²)......(3)

[Rearranging the terms]


ΔABC is a right triangle,

AB²+BC² = AC²…………….(4)
(by the Pythagoras theorem)



Δ DBE is a right triangle

DB²+BE² = DE²………………(5)

(by the Pythagoras theorem)



AE²+CD²= (AB²+BC²)+(BE²+BD²)


AE²+CD²= AC²+DE²

[ From equation 4 and 5]

[FIGURE IS THE ATTACHMENT FIGURE IS THE ATTACHMENT]

Answer 2

ΔABC is a right triangle, so by the Pythagorean

theorem,

(1) AB²+BC² = AC²

ΔABE is a right triangle, so by the Pythagorean

theorem,

(2) AB²+BE² = AE².

ΔDBC is a right triangle, so by the Pythagorean

theorem,

(3) DB²+BC² = CD².

ΔDBE is a right triangle, so by the Pythagorean

theorem,

(4) DB²+BE² = DE².

Add BE²+DB² to both sides of equation (1)

AB²+BC²+BE²+DB² = AC²+BE²+DB²

Rearrange the terms:

AB²+BE²+DB²+BC² = AC²+DB²+BE²

To make things easier to see, let's put parentheses

around the first two terms on the left, the last two

terms on the left, and the last two terms on the right:

(5) (AB²+BE²)+(DB²+BC²) = AC²+(DB²+BE²)

Using (2) we replace (AB²+BE²) in (5) by AE²

Using (3) we replace (DB²+BC²) in (5) by CD²

Using (4) we replace (DB²+BE²) in (5) by DE²

And we end up with what we were to prove:

AE²+CD² = AC²+DE²