Question 3

In a triangle $$ABC$$, $$a$$, $$b$$, $$c$$ denote the lengths of the sides $$BC$$, $$CA$$, $$AB$$. If $$D$$ is the midpoint of the side $$BC$$ and $$AD$$ is perpendicular to $$AC$$, then

Solution

Triangle $$ADC$$ is right angled at $$A$$, so $$\left(\frac{a}{2}\right)^2 = AD^2 + b^2$$. By Apollonius' theorem $$AD^2 = \frac{2b^2 + 2c^2 - a^2}{4}$$. Substituting, $$a^2 = 2b^2 + 2c^2 - a^2 + 4b^2$$, i.e. $$a^2 = 3b^2 + c^2$$, so $$3b^2 = a^2 - c^2$$.

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