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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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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.
Solved.