Question 12In a quadrilateral ABCD- - A - - D-90- Prove that AC2-BD2-AD2-BC2-

Given, quadrilateral ABCD, in which ∠ A+∠ D=90^∘ nbsp; To prove : AC^2+BD^2=AD^2+BC^2 Construction : Produce AB and CD to meet at E. Also, join AC and BD. ...

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Standard VII

Mathematics

Right-angled Triangle

Question 12In...

Question

Question 12
In a quadrilateral ABCD, $∠ A + ∠ D = 90^{\circ}$ . Prove that $A C^{2} + B D^{2} = A D^{2} + B C^{2}$ .

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Solution

Given, quadrilateral ABCD, in which
$∠ A + ∠ D = 90^{\circ}$

To prove :
$A C^{2} + B D^{2} = A D^{2} + B C^{2}$

Construction : Produce AB and CD to meet at E.
Also, join AC and BD.

Proof :
In $Δ A E D$ ,
$∠ A + ∠ D = 90^{\circ} [g i v e n]$
$∴ ∠ E = 180^{\circ} - (∠ A + ∠ D) = 90^{\circ}$ $[b e c a u s e s u m o f a l l a n g l e s i n a t r i a n g l e = 180^{\circ}]$

Then, by using Pythagoras theorem,
$A D^{2} = A E^{2} + D E^{2} . . (i)$

$I n Δ B E C$ ,
by using Pythagoras theorem,
$B C^{2} = B E^{2} + E C^{2} . . . (i i)$

On adding equations (i) and (ii), we get,
$A D^{2} + B C^{2} = A E^{2} + D E^{2} + B E^{2} + C E^{2} . . . (i i i)$

$I n Δ A E C$ ,
by using Pythagoras theorem,
$A C^{2} = A E^{2} + C E^{2} . . . (i v)$

$& i n D e l t a B E D$ ,
by using Pythagoras theorem,
$B D^{2} = B E^{2} + D E^{2} . . . (v)$

On adding both equations, we get,
$A C^{2} + B D^{2} = A E^{2} + C E^{2} + B E^{2} + D E^{2} . . . (v i)$

From Eq. (iii) and (vi), we get,
$A C^{2} + B D^{2} = A D^{2} + B C^{2}$

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Q. In a quadrilateral ABCD,

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In a quadrilateral ABCD, $∠ B = 90^{\circ}, A D^{2} = A B^{2} + B C^{2} + C D^{2}$ , prove that $∠ A C D = 90^{\circ}$

In a trapezium ABCD, if AB || CD then $(A C^{2} + B D^{2})$ = ?

(a) $B C^{2} + A D^{2} + 2 B C . A D$

(b) $A B^{2} + C D^{2} + 2 A B . C D$

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Q. In a trapezium ABCD, if AB || CD, then (AC² + BD²) = ?
(a) BC² + AD² + 2BC ⋅ AD
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Question 12 In a quadrilateral ABCD, ∠A+∠D=90∘. Prove that AC2+BD2=AD2+BC2.

Question 12
In a quadrilateral ABCD, $∠ A + ∠ D = 90^{\circ}$ . Prove that $A C^{2} + B D^{2} = A D^{2} + B C^{2}$ .