Question

10. In the adjoining figure, AD is the angle bisector of exterior Z A of A ABC.
BD bisects Z B. If AB = AC, then prove that
(i) AD || BC
(ii) AB = AD.​

Answer 1

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$\sf \: given : abc \: is \: a \: triangle \\ \sf \: ad \: is \: the \: exterior \: bisector \: of \angle \: a \: and \: meets \: bc \\ \\ \sf produced \: at \: d \: and \: ba \: is \: produced \: to \: f\: \\ \green \bigstar \ \bf \: to \: prove : \frac{bd}{cd} = \frac{ab}{ac} \\ \angle \: cad = \angle \: ace \: (alternate \: angles) \\ similarly \: ce \parallel \: ad \: cut \: by \: ab \\ \angle \: fad = \angle \: aec \: (corresponding \: angles) \\ ac = ae(by \: isosceles \red\triangle \: theorem) \\ now \: in \green \triangle \: bad \:. \: ce \parallel \: da \\ ae = dc(basic \: proportionality \: theorem) \\ \bf \: ab = bd \: \\ but \: ac = ae(proved \: above) \\ ac = dc \\ \: \bf ab = bd \: or \: \frac{ab}{ac} = \frac{bd}{dc} (hence \: proved)$