SOLUTION: ABC is an isosceles triangle in which mAB =mAC.CD is the perpendicular drawn from C to the opposite side .Prove that (mBC)^2=2 (mAB).(mBD).

1.  Make a sketch.
    Draw the triangle ABC and the perpendicular CD.

    Draw the median AE from the vertex A to the base BC.
    Notice that the median AE is the altitude of the triangle ABC at the same time.


2.  The right-angle triangles BCD and ABE are SIMILAR.
        ( Because they have the common acute angle B. 
          For right-angled triangles having a common acute angle, it is enough to be similar ! )


3.  From the similarity, you have this proportion for the measures of corresponding sides:

     = .      (1)


4.  Now notice that BE = .

    Together with (1), it implies that  = 2*|BD|*|AB|. 

    QED.