Try this beautiful problem from PRMO, 2018 based on Triangles.
In the triangle, ABC, the right angle at A. The altitude through A and the internal bisector of \(\angle A\) have lengths \(3\) and \(4\), respectively. Find the length of the medium through A?
Geometry
Triangle
Pythagoras
Answer:\(24\)
PRMO-2018, Problem 12
Pre College Mathematics
We have to find out the value of \(AM\) .Now we can find out the area of Triangle dividing two parts , area of \(\triangle AMC\) + area of \(\triangle ABM\) ( as M is the mid point of BC)
Can you now finish the problem ..........
Now \(AN\) is the internal bisector...Therefore \(\angle NAB=\angle NAC= 45^{\circ}\).Let \(AC=b\) ,\(AB=c\) and \(BC=c\).using this values find out the area of triangles AMC and Triangle ABM
Area of \(\triangle ABC\)=\(\frac{1}{2} bc=\frac{1}{2} \times a \times 3\)
\(\Rightarrow bc=3a\)......................(1)
Now Area of \(\triangle ABN\) + Area of \(\triangle ANC\)=Area of \(\triangle ABC\)
\(\Rightarrow \frac{1}{2} c 4 sin 45^{\circ} +\frac{1}{2} b 4 sin 45^{\circ}=\frac{1}{2} bc\)
\(\Rightarrow b+c =\frac{1}{2\sqrt 2} bc\)
(squarring both sides we get..........)
\(\Rightarrow b^2 +c^2 +2bc=\frac{1}{8} b^2 c^2\)
\(\Rightarrow a^2 +6a=\frac{9}{8} a^2\) (from 1)
\(\Rightarrow a +6 =\frac{9}{8} a\) \((as a \neq 0)\)
\(\Rightarrow a=48\)
\(\Rightarrow AM=BM=MC=\frac{a}{2}=24\)
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