In a trapezium $A B C D$, if $A B \| C D$, then $\left(A C^{2}+B D^{2}\right)=?$
(a) $B C^{2}+A D^{2}+2 B C \cdot A D$
(b) $A B^{2}+C D^{2}+2 A B \cdot C D$
(c) $A B^{2}+C D^{2}+2 A D \cdot B C$
(d) $B C^{2}+A D^{2}+2 A B \cdot C D$
(c) $B C^{2}+A D^{2}+2 A B \cdot C D$
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.
$\therefore A C^{2}=B C^{2}+A B^{2}-2 A B \cdot A E$
In ∆ABD, ∠A is acute.
$\therefore B D^{2}=A D^{2}+A B^{2}-2 A B \cdot A F$
$\therefore A C^{2}+B D^{2}=\left(B C^{2}+A D^{2}\right)+\left(A B^{2}+A B^{2}\right)-2 A B(A E+B F)$
$=\left(B C^{2}+A D^{2}\right)+2 A B(A B-A E-B F) \quad[\because A B=A E+E F+F B$ and $A B-A E=B E]$
$=\left(B C^{2}+A D^{2}\right)+2 A B(B E-B F)$
$=\left(B C^{2}+A D^{2}\right)+2 A B \cdot E F$
$\therefore A C^{2}+B D^{2}=\left(B C^{2}+A D^{2}\right)+2 A B . C D$