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  • ABCD is a quadrilateral and E and F are points on AD and CD respectively such that AB = CB, <img alt="\angle ABE =\angle CBF" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cangle%20ABE%20%

ABCD is a quadrilateral and E and F are points on AD and CD respectively such that AB = CB, \angle ABE =\angle CBF and \angle EBD =\angle FBD.

Option: 1

BE=BF


Option: 2

BE>BF


Option: 3

BE<BF


Option: 4

None of the above


Answers (1)

best_answer

\begin{aligned} &\angle A B E=\angle C B F \text { and } \angle E B D=\angle F B D\\ &\Rightarrow \angle A B E+\angle E B D=\angle C B F+\angle F B D\\ &\Rightarrow \angle A B D = \angle C B D \end{aligned}

\\\text{Now, in } \triangle A B D\text{ and }\triangle C B D,\text{ we have} \\ \\A B=C B \quad(\text { given }) \\\\\angle A B D=\angle C B D \qquad {[\text{ from (i) }]} \\ \\B D=B D \text{ ( common ) } \\\\\therefore \quad \triangle A B D \cong \triangle C B D \quad[ \text { SAS-criteria] } \\\\ \therefore \angle B A D=\angle B C D \quad[\text { c.p.c.t. }]

\begin{align*} &\Rightarrow \angle BAE = \angle BCF \\ &\text{Now, in } \triangle ABE \text{ and } \triangle CBF, \text{ we have:} \\ &AB = CB \quad \text{(given)} \\ &\angle ABE = \angle CBF \quad \text{(given)} \\ &\angle BAE = \angle BCF \quad \text{[proved in (ii)]} \\ \therefore &\triangle ABE \cong \triangle CBF \quad \text{(by AAS-criteria)} \\ \text{Hence, } &BE = BF. \end{align*}

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Pankaj Sanodiya

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