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ABC is an isosceles triangle such that AB = AC and angle B = 35 degree, AD is median to base BC , then angle BAD is ?

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by debanti

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by debanti

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Let the angle ADB be x

then ADC is 180-x

as AD is median so BD=CD

and for isosceles triangle AB=AC

so AB/AC=BD/CD=1

By angle bisector theorem

BAD=CAD=y (just take it)

for an triangle BAD 35+x+y=180...(1)

for an triangle DAC 35+180-x+y=180...(2)

35+y=x

therefore

35+34+y+y=180

2y+70=180

2y=100

y=55

therefore angle BAD=CAD=55

The Brainliest Answer!

then angle A will be 180-35-35 since sum of angles in a triangle is 180 degrees.hence,angle A is 110 degrees.

angle BAD is 110/2 degrees as 'in isosceles triangle,median drawn to base is it's altitude and angular bisector'.....

hence,angle BAD=55 degrees

hope this is correct and helps u......