Question

In the given figure,ABCD is a quadrilateral and tnf are points on a d and c respectively such that AB=CB,∠ABE=∠CBF and ∠EBD=∠FBD.Prove that BE=BF.​

Answer 1

Answer:

PLs SENd

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Answer 2

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In the given figure,ABCD is a quadrilateral and tnf are points on a d and c respectively such that AB=CB,∠ABE=∠CBF and ∠EBD=∠FBD.Prove that BE=BF.

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$\angle \: ABE = \angle \: CBF \: and \: \angle \: EBD = \angle \: FBD$

$⇒ \angle \: ABE + \angle \: EBD = \angle \: CBF + \angle \: FBD$

$⇒ \angle \: ABD = \angle \: CBD \: \: \: \: \: \: ..(i)$

Now, in ∆ABD and ∆CBD,we have

$AB = CB \: \: \: \: (given)$

$\angle \: ABD = \angle \: CBD \: \: \: (from \: (i))$

$BD = BD \: \: \: \: \: \: \: \: \: (common)$

$\therefore \: \triangle \: ABD \: \cong \: \triangle \: CBD \: \: \: (sas - criteria)$

$\therefore \: \angle \: BAD = \angle \: BCD \: \: \: (c.p.c.t).$

$⇒ \angle \: BAE = \angle \: BCF. \: \: \: \: \: \: ...(ii)$

Now, in ∆ABE and ∆CBF,we have

$AB = CB \: \: \: (given)$

$\angle \: ARE = \angle \: CBF \: \: \: (given)$

$\angle \: BAE = \angle \: BCF \: \: \: [proved \: in \: (ii)]$

$\therefore\triangle \: ABE \: \cong \: \triangle \: CBF \: \: \: (aas - criteria)$

Hence,BE = BF.

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