In a quadrilateral ABCD - - A - D -90- Prove that AC 2- BD 2- AD 2- BC 2

The correct option is Given Qusdrilateral ABCD, in which ∠ A+∠ D=90^∘ To prove AC^2+BD^2=AD^2+BC^2 Construct Produce AB and CD to meet at E. Also, join AC ...

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Standard VIII

Mathematics

Rectangle

In a quadrila...

Question

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 Qusdrilateral ABCD, in which $∠ A + ∠ D = 90^{\circ}$
To prove $A C^{2} + B D^{2} = A D^{2} + B C^{2}$
Construct Produce AB and CD to meet at E.
Also, join AC and BD.
Proof In $Δ A E D$ , $∠ A + ∠ D = 90^{\circ}$ [given]
$∴$ $∠ E = 180^{\circ} - (∠ A + ∠ D) = 90^{\circ}$ $[∴ s u m o f a n g l e s o f a t r i a n g l e = 180^{\circ}]$
Then, by Pythagoras theorem, $A D^{2} = A E^{2} + D E^{2}$
In $Δ B E C$ , by Pyhtagoras theorem, $B C^{2} = B E^{2} + E F^{2}$
On adding both equations, we get
$A D^{2} + B C^{2} = A E^{2} + D E^{2} + B E^{2} + C E^{2}$ ……(i)
In $Δ A E C$ by Pythagoras theorem,
$A C^{2} = A E^{2} + C E^{2}$
And in $Δ B E D$ , by Pythagoras theorem,
$B D^{2} = B E^{2} + D E^{2}$
On adding both equations, we get
$A C^{2} + B D^{2} = A E^{2} + C E^{2} + B E^{2} + D E^{2}$
From Eq. (i) and (ii),
$A C^{2} + B D^{2} = A D^{2} + B C^{2}$

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Q. Question 12
In a quadrilateral ABCD,

∠ A + ∠ D = 90^{\circ}

. Prove that

A C^{2} + B D^{2} = A D^{2} + B C^{2}

Q. In a quadrilateral ABCD

∠ A + ∠ D = 90^{0}

, then prove that

B D^{2} + A C^{2} = A D^{2} + B C^{2}

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

Q. In a quadrilateral

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{A D}^{2} = {A B}^{2} + {B C}^{2} + {C D}^{2}

, prove that

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

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

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In a quadrilateral ABCD, ∠A+∠D=90∘ Prove that AC2+BD2=AD2+BC2

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