[tex]\large\bf{\underline{Given :}}[/tex]
- An isosceles triangle ABC
- AD is a bisector of ∠A
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[tex]\large\bf{\underline{To \:prove :}}[/tex]
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[tex]\large\bf{\underline{In \: triangle \: ABD\:and\: triangle\:ADC}}[/tex]
[tex]{⟹AD = AD\:\:\:\:\:\:\:\:\:\:\:\:\:\:( common)}[/tex]
[tex]{⟹∠BAD = ∠CAD\:\:\:\:\:\:\:\:\:( given : AD \:is \:bisector \:of \:∠A )}[/tex]
[tex]{⟹∠ABD = ∠ACD\:\:\:\:\:\:\:\:\:( given : angles\: in \:isosceles\: trianlge )}[/tex]
[tex]\large\bf{\underline{Hence}}[/tex]
- By ASA criteria ∆ABD congurant to ∆ADC
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[tex]\large\bf{\underline{So}}[/tex]
- By CPCT ( corresponding parts of congurant triangle )
[tex]\large\bf{BD = DC}[/tex]
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[tex]\large\bf{\underline{Note :}}[/tex]Slide to view full answer.
