Question

in the given figure, AD = AB and AE bisects angle A.Prove that:i) BE = ED ii)angleABD >angle BCA

Answer 1

$\underline{\pink{ Given :}}$

$In \: the \: given \: figure , AD = AB \:and$

$AE \: bisects \: \angle A$

$\underline { \blue{ Solution : }}$

$i) In \: \triangle ABE \:and \: \triangle ADE$

$AB = AD \: ( given )$

$\angle {BAE} = \angle {DAE}$

$AE = AE \: ( Common \: side )$

$\therefore \triangle ABE \cong \triangle ADE$

$\blue { ( S.A.S \: congruence \: rule ) }$

$BE = ED \: ( C.P.C.T )$

$ii) In \: \triangle BCD, \: CD \: extended \: to \:A$

$\angle {BDA } \gt \angle {BCA}$

$\blue {( \because Exterior \:angle \: at \: D \:is}$

$\blue { is \: greater \:than \: interior\: opposite}$

$\blue { angle ) }$

$\implies \angle {ABD } \gt \angle {BCA}$

$\blue{ ( \because \angle {BDA} = \angle {ABD }) }$

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