Question

Prove that in an isosceles triangle, the median to the base is also perpendicular.

Answer 1

R.E.F image

Let AD be the median

of $△$ ABC

$\therefore$ In $△$ ABD & $△ADC$

AB=AC[isosceles triangle]

BD=DC [median]

$\angle ABD=\angle ACD$ [ angle of isosceles triangle]

$\therefore △ABD=△ADCso\angle ADB=\angle ADB$

$so\angle ADB+\angle ADC={180}^{\circ }$ [liner angle]

$so\angle ADC={90}^{\circ }$[median is perpendicular]

Hence proved.