# A problem of a triangle

Consider a triangle $ABC$ where the median $CM$ is perpendicular to the angle bisector $AL$ and their ratio is $\sqrt2 : 1$. The question is to find $\cos A$. Hints?

Btw, I do know that the triangle $AMC$ is isosceles.

• Does $AMC$ iscosceles mean that $AM=MC?$ – Igor Rivin Dec 4 '13 at 17:54
• No, I meant $AM = AC$ – Parth Thakkar Dec 4 '13 at 17:55

As you said, $AL$ is both altitude and bisector in $AMC$ which shows that $AC=AM$.

Hence

$$AB=2AC$$

Let $BC=a, AC=b$ then $AB=2b$.

Using the length of the bisector and median you have

$$CM^2=\frac{b^2+a^2}{2}-\frac{4b^2}{4}=\frac{a^2-b^2}{2}$$ $$AL^2=\frac{4b2b}{(b+2b)^2}[(b+2b)^2-a^2]=\frac{8}{9}(9b^2-a^2)$$

Now, plugging in $\frac{CM^2}{AL^2}=2$ you get a linear equation in $\frac{a^2}{b^2}$. Solve and use

$$\cos(A)=\frac{b^2+4b^2-a^2}{4b^2}$$

• I was unaware of these formulas (median and bisector lengths). – Parth Thakkar Dec 4 '13 at 18:04

Here is a simpler geometric proof.

Extend $AC$ by $CC'=AC$.

Then, by symmetry, $C,L M$ are colinear and $AM=MB=AC=CC'$.

As $CM$ is half line, $CM=\frac{BC'}{2}$.

As $\Delta LMC \sim \Delta LBC'$ we have $\frac{DL}{D'L}=\frac{MC}{BC'}=\frac{1}{2}$ and hence $\frac{DL}{DD'}=\frac{1}{3}$.

As $MC$ is a half line, we also have $DD'=AD$. Thus

$$\frac{DL}{AD}=\frac{1}{3} \,.$$ From here we get

$$\frac{AD}{AL}=\frac{3}{4}$$

Then

$$\frac{AD}{DC}=2 \frac{AD}{MC}=2 \frac{AD}{AL} \frac{AL}{MC}=2 \cdot \frac{3}{4} \frac{1}{\sqrt{2}}=\frac{3 \sqrt{2}}{4}$$

Now, you can easely find $\sin(\frac{A}{2})$ and $\cos(\frac{A}{2})$ in $\Delta ADC$ and use double angle formula.