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

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

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

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Solution

ANSWER:
(c) $B C^{2} + A D^{2} + 2 A B . C D$

Explanation:
Draw perpendicular from D and C on AB which meets AB at E and F, respectively.
∴ DEFC is a parallelogram and EF = CD .
In ∆ ABC, ∠ B is acute.
$∴ A C^{2} = B C^{2} + A B^{2} - 2 A B . A E$
In ∆ ABD, ∠ A is acute.
∴
$B D^{2} = A D^{2} + A B^{2} - 2 A B . A F$
∴ $A C^{2} + B D^{2} = (B C^{2} + A D^{2}) + (A B^{2} + A B^{2}) - 2 A B (A E + B F)$
= $(B C^{2} + A D^{2}) + 2 A B (A B - A E - B F)$
[∵ AB = AE + EF + FB and AB - AE = BE ]
= $(B C^{2} + A D^{2}) + 2 A B (B E - B F)$
= $(B C 2 + A D 2) + 2 A B . E F$
$∴ A C^{2} + B D^{2} = (B C^{2} + A D^{2}) + 2 A B . C D$

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