Question

1. In the fig. AC = AE and AB = AD and ZBAD = ZEAC. Prove that BC = DE.
(C.P.C.T.) Proved.
Join DE
ZBAD = ZEAC
$2 \sqrt[ \tan( \sec(\pi \alpha \pi \binom{\%\pi\pi\pi \beta \cos( \sin( \sec( \alpha \gamma \sec( \sec( log(\pie\pi ln(\pi ln(\pi\pi \tan( log_{eee\pi\pi\pi\pi\pi}(?) ) ) ) ) ) ) ) ) ) }{?} ) ) ]{?}$
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Answer 1

Answer:

\sec( log(\pie\pi ln(\pi ln(\pi\pi \tan( log_{eee\pi\pi\pi\pi\pi}(?) ) ) ) ) ) ) ) ) ) }{?} ) ) ]{?} [/tex]