ABCD is a trapezium in which $A B ∥ C D$ . Then ${A C}^{2} + {B D}^{2}$ is equal to:

{A D}^{2} + {B C}^{2} - 2 A B . C D

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

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{A D}^{2} - {B C}^{2} + 2 A B . C D

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{A D}^{2} - {B C}^{2} - 2 A B . C D

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Solution

The correct option is B ${A D}^{2} + {B C}^{2} + 2 A B . C D$
Using Pythagoras Theorem
$A C^{2} + B D^{2}$
$= (A L^{2} + L C^{2}) + (B M^{2} + M D^{2})$
$= (A M + M L)^{2} + L C^{2} + (B L + L M)^{2} + M D^{2}$
$= (A M^{2} + M D^{2}) + (B L^{2} + L C^{2}) + 2 M L^{2} + 2 A M . M L + 2 B L . M L$
$= A D^{2} + B C^{2} + 2 M L . (M L + A M + B L)$
$= A D^{2} + B C^{2} + 2 C D . A B$ $(M L = C D}$

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Similar questions

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

∠ A + ∠ D = 90^{0}

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

∠ A + ∠ D = 90^{\circ}

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

∠ A + ∠ D = 90^{\circ}

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

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