Question

1. In figure A ABC is right angled with ZB = 90°
Prove that AE? + CD2 = AC? + DE​

Answer 1

QUESTION

In the figure Δ ABC is right angled with ∠ B = 90°

Prove that  AE + CD² = AC² + DE²

PROOF

In Δ ABC

by

AC² = AB² + BC² ------equation(1)

In Δ DBE

by Pythagoras

DE² = BD² + BE² ------equation (2)

adding equation (1) and (2)

AC² + DE² = AB² + BC² + BD² + BE²

AC² + DE² = (AB² + BE² ) + ( BC² + BD² )

( by pythagoras in Δ ABE we have

AB² + BE² = AE²

and by pythagoras in Δ DCB we have

BC² + BD² = CD² )

so,

AC² + DE² = AE² + CD²      

PROVED.

Answer 2

HEY MATE YOUR ANSWER IS HERE...

★ CORRECT QUESTION ★

:- IN FIGURE,TRIANGLE ABC IS A RIGHT ANGLE TRIANGLE , RIGHT ANGLED AT B THEN ,

PROVE THAT AE² + CD² = AC² + DE²

★ GIVEN ★

:- TRIANGLE ABC IS A RIGHT ANGLE TRIANGLE, RIGHT ANGLED AT B

★ TO PROVE ★

:- AE² + CD² = AC² + DE²

★ PROOF ★

IN TRIANGLE ABE

ANGLE B = 90° ( GIVEN )

HENCE,

BY PHYTOGORAS THEOREAM

AE² = AB² + BE² --------Eq 1

IN TRIANGLE BDC

ANGLE B = 90° ( GIVEN )

HENCE,

BY PHYTOGORAS THEOREAM

CD² = DB² + BC² --------Eq 2

NOW BY ADDING EQUATIONS 1 & 2

AE² + CD² = AB² + BE² + DB² + BC²

BY REARRANGING THE EQUATION

AE² + CD² =

( AB² + BC² ) + ( BE² + DB² ) --------Eq 3

$now \:in \: triangle \: </strong><strong>ABC</strong><strong> \\ \: angle \: </strong><strong>B</strong><strong> = 90 \: degree \: (given) \\ \\ hence \: by \: \\ PHYTOGORAS \: THEOREAM \: \: \: \: \\ \\ {</strong><strong>AC</strong><strong>}^{2} = {</strong><strong>AB</strong><strong>}^{2} + {</strong><strong>BC</strong><strong>}^{2} \: - - - \: \: eq \: 4$

SIMILARLY,

$now \:in \: triangle \: </strong><strong>BED</strong><strong> \\ \: angle \: </strong><strong>B</strong><strong> = 90 \: degree \: (given) \\ \\ hence \: by \: \\ PHYTOGORAS \: THEOREAM \: \: \: \: \\ \\ {</strong><strong>DE</strong><strong>}^{2} = {</strong><strong>BD</strong><strong>}^{2} + {</strong><strong>BE</strong><strong>}^{2} \: - - - \: \: eq \: 5$

NOW PUT EQUATIONS 4 & 5 IN EQUATION 3

AE² + CD² = AC² + DE²

HENCE PROVED

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