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For an isosceles triangle angles ABC=ACB=35 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

As angle B is 35 degree angle C is also 35 degrees because angles opp. to equal sides are equal. 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......

friends plase tell me the clear meaning of the statements , "angle BAD is 110/2 degrees as 'in isosceles triangle,median drawn to base is it's altitude and angular bisector'....."

In isosceles triangle,median drawn to base is same as it's altitude and angular bisector.so,by taking angular bisector I hv halved angle A into half...ok...?