Question

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​

Answer 1

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

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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

Answer 2

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