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ad is an altitude of an isosceles triangle ABC in which AB equal to AC show that ad bisects BC and AD bisects angle A |
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Answer» Maths Ad is an altitude of an isosceles triangle ABC in which AB=AC.Show that AD bisects BC. December 20, 2019avatar Hitesh Rajmohan SHARE ANSWER Given that, \triangle ABC△ABC is an isosceles triangle so, AB=AC--(i)AB=AC−−(i) ALSO, AD is the altitude so, \ANGLE ADC=\angle ADB=90---(ii)∠ADC=∠ADB=90−−−(ii) To prove : (i) BD=CD(i)BD=CD (ii)\angle BAD=\angle CAD(ii)∠BAD=∠CAD Proof :- In \triangle ADB△ADB and \triangle ADC△ADC \angle ADC=\angle ADB=90∠ADC=∠ADB=90 AB=ACAB=AC[from (i)(i)] AD=ADAD=AD \triangle ADB\cong \triangle ADC△ADB≅△ADC Hence,by CPCT BD=DCBD=DC and \angle ABC=\angle DAC∠ABC=∠DAC hence,proved |
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