Question 1067733
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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).
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<pre>
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:

    {{{abs(BD)/abs(BC)}}} = {{{abs(BE)/abs(AB)}}}.      (1)


4.  Now notice that BE = {{{(1/2)*abs(BC)}}}.

    Together with (1), it implies that {{{abs(BC)^2}}} = 2*|BD|*|AB|. 

    QED.
</pre>

Solved.