Medians | Geometry | PRMO-2018 | Problem 13

In a triangle ABC, right-angled at A, the altitude through A and the internal bisector of \(\angle A \) have lengths \(3\) and \(4\), respectively. Find the length of the median through \(A\).

Now in the Right angle Triangle ABC ,\(\angle A=90\).\(AD=3\) is Altitude and \(AE=4\) is the internal Bisector and \(AF\) is the median.Now we have to find out the length of \(AF\)

Now \(\angle CAE = 45^{\circ }= \angle BAE\).

Let\( BC = a\), \(CA = b\), \(AB = c\)
so \(\frac{bc}{2}=\frac{3a}{2}\) \(\Rightarrow bc=3a\)

Can you now finish the problem ..........

Therefore,

\(\frac{2bc}{b+c} cos\frac{A}{2}=4\)

\(\Rightarrow \frac{6a}{b+c} .\frac{1}{\sqrt 2}=4\)

\(\Rightarrow 2\sqrt 2 (b+c)=3a\)

\(\Rightarrow 8(b^2 + c^2 + 2bc) = 9a^2\)

\(\Rightarrow 8(a^2 + 6a) = 9a^2\)

\(\Rightarrow 48a=a^2\)

\(\Rightarrow a=48\)

Can you finish the problem........

Now \(AF\) is the median , \(AF=\frac{a}{2}=24\)

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