Question

Pls help and no unnecessary typin* I will report

Answer 1

$\large\bf{\underline{Given :}}$

• An isosceles triangle ABC
• AD is a bisector of ∠‎A

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$\large\bf{\underline{To \:prove :}}$

• BD = DC

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$\large\bf{\underline{In \: triangle \: ABD\:and\: triangle\:ADC}}$

${⟹AD = AD\:\:\:\:\:\:\:\:\:\:\:\:\:\:( common)}$

${⟹∠BAD = ∠CAD\:\:\:\:\:\:\:\:\:( given : AD \:is \:bisector \:of \:∠A )}$

${⟹∠ABD = ∠ACD\:\:\:\:\:\:\:\:\:( given : angles\: in \:isosceles\: trianlge )}$

$\large\bf{\underline{Hence}}$

• By ASA criteria ∆ABD congurant to ∆ADC

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$\large\bf{\underline{So}}$

• By CPCT ( corresponding parts of congurant triangle )

$\large\bf{BD = DC}$

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$\large\bf{\underline{Note :}}$Slide to view full answer.