Question

ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2 AD.

Answer 1

ANSWER✔

$\large\underline\bold{GIVEN,}$

$\dashrightarrow \triangle BAC \:is\:a\:right\:angled\:triangle.$

$\dashrightarrow \angle BAC=90\degree$

$\dashrightarrow AB= AC$

$\dashrightarrow line\:AD\:bisects\:BC \\ \angle BAD= \angle DAC= \dfrac{1}{2} \angle BAC\\ \angle BAD= \angle DAC = \dfrac{90}{2}\\ \cancel\dfrac{90}{2} \\ \boxed{\angle BAD= \angle DAC =45\degree}$

$\large\underline\bold{TI\:PROVE,}$

$\dashrightarrow 2AD=BC$

$\large\underline\bold{SOLUTION,}$

$\dashrightarrow AB=AC \\ \therefore \angle B = \angle C$

$\therefore by\:angle\:sum\:property\:of\:triangle$

$\dashrightarrow \angle A +\angle B +\angle C=180\degree$

$\implies \angle A +\angle B +\angle B = 180\degree\:--\boxed{\angle B = \angle C }$

$\implies 90+ 2\angle B =180$

$\implies 2 \angle B = 180-90$

$\implies 2\angle B=90$

$\implies \angle B= \dfrac{90}{2}$

$\implies \angle B= \cancel\dfrac{90}{2}$

$\implies \angle B=45$

NOW,

$\therefore in\:\triangle ABD \: and \: \triangle ACD$

$\dashrightarrow AD=AD \:--- common\:side.$

$\dashrightarrow BD= CD \:--- as\: AD \:bisect\:BC$

$\dashrightarrow \angle B= \angle C \:---each\:45\degree$

$\sf\therefore \triangle ABD \cong \triangle ACD\:--- SAS\:RULE\:of\:congruency.$

$\dashrightarrow \angle ADB = \angle ADC \:---90\degree\:---c.p.c.t.$

THEREFORE

$\dashrightarrow in\: \triangle ABD \\ \therefore\: by\: Pythagoras\: theorem,\\ AB^2=BD^2+AD^2\:---\boxed{1}$

$\dashrightarrow in\: \triangle ADC \\ \therefore\: by\: Pythagoras\: theorem,\\ AC^2=CD^2+AD^2\:---\boxed{2}$

$\therefore now,adding\:these\:both\:we\:get\:,$

AB²=BD²+AD²

AC²=CD²+AD²

————————————

AB²+AC²=2AD²+ BD²+CD²

$\implies AB^2+AC^2=2AD^2+ 2BD^2\:---\boxed{BD=CD}$

$\implies AB^2 +AC^2= 2(AD^2+BD^2)$

$\implies BC^2= 2(AD^2+BD^2)\:--\boxed{AB+AC=BC}$

$\implies \dfrac{BC^2}{2} =AD^2 + \bigg( \dfrac{ BC}{2} \bigg) \:---\boxed{ BD = \dfrac{1}{2} BC}$

$\implies \dfrac{BC^2}{2} + \dfrac{BC^2}{4}= AD^2$

$\implies \dfrac{4BC^2-2BC^2}{8}= AD^2$

$\implies \dfrac{2BC^2}{8}= AD^2$

$\implies \dfrac{\cancel{2}\:BC^2}{\cancel{8}}= AD^2$

$\implies \dfrac{BC^2 }{2} = AD^2$

$\implies BC^2=2AD^2$

$\implies BC^{\cancel{2}}=2AD^{\cancel{2}}$

$\implies BC=2AD$

$\large{\boxed{\bf{ \star\:\: HENCE\:PROVED\:\: \star}}}$

__________________

$\red{\text{FOR DIAGRAM PLEASE REFER THE ATTACHMENT.}}$

Answer 2

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

$ABC \:is \: a \: right \: triangle .$

$AB = AC \: ---(1)$

$and \: Bisector \: of \: \angle A \: meets \:BC$

$at \: D$

$\underline{\pink{ To \: Prove :}}$

$\blue {BC = 2AD }$

$\underline{\pink{ Proof :}}$

$i) In \: \triangle ABC, \angle A = 90\degree$

$AB = AC \: (given )$

$\angle { ABC } = \angle { ACB}$

$\blue{ ( \because Angles \: opposite \: to }$

$\blue { equal \: sides \: are \; equal ) }$

$\therefore \angle { ABC } = \angle { ACB} = 45\degree \: --(2)$

$ii) Bisector \: of \: \angle A \: meets \:BC$

$\angle { BAD } = \angle {ABD} = 45\degree$

$AD = BD \: --(3)$

$\blue{ ( \because Angles \: opposite \: to }$

$\blue { equal \: sides \: are \; equal ) }$

$iii) Similarly,$

$\angle { ACD } = \angle {DAC} = 45\degree$

$AD = DC \: --(4)$

/* Add Equations (3) and (4) , we get */

$AD + AD = BD + DC$

$\implies 2AD = BC$

$Hence\: proved$

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