Draw a diagram and prove that, In an isosceles triangle, the median joining the vertex formed by intersection of equal sides) to the mid point of opposite side is also an altitude.

Hi,
Lets prove in reverse order that if mentioned condition is true then it will be an isosceles triangle.
We form our diagram from given information , As :

Here AD is our median and perpendicular on line BC , So

BD  =  CD                                                                        ---- ( 1 )

and

∠ ADB  =  ∠ ADC  =  90°                                 ---- ( 2 )

In ∆ ADB  and ∆ ADC 

BD  =  CD                                                                       ( From equation 1 )

∠ ADB  =  ∠ ADC                                                 ( From equation 2 )

And

AD  =  AD                                                                       ( Common side )

Therefore ,

∆ ADB  and ∆ ADC                                                 ( By SAS rule )

So,

AB  =  AC                                                                         ( By CPCT )

In ∆ ABC , AB  =  AC ( As we proved ) , So

∆ ABC is a isosceles triangle .                                        ( Hence proved )
Regards