Right triangle with a altitude and a median line Can anyone help me on this? I have hard time to find the solution. Thank you very much!
In $\triangle ABC$ with right angle at $C$, altitude $CH$ and median $CM$ trisect the right angle. If the area of $\triangle CHM$ is $K$, find the area of $\triangle ABC$.

 A: It is a classical construction using four times a half equilateral triangle (see figure). Indeed, consider the equilateral triangle $AMC$, taking $AM=AC=1$,we have $MB=AC=AM$, proving that $M$ is the midpoint of $AB$, with each angle in $C$ equal to $\pi/6$.
Thus the answer is that the area of ABC is $4K$, i.e., four times the area of CHM. 

A: If the right angle is trisected, then angle ACH measures 30 degrees and angle ACM measures 60 degrees. You are to prove that triangle CHM is similar to triangle ABC and work out the ratio between corresponding side lengths.  The area of triangle ABC then follows.
A: Observe that 
$$\;\Delta ABC\sim\Delta ACH\sim\Delta CBH\sim \Delta MCH\;$$
and also that
$$\;\angle BCM=\angle MCH=\angle HCA=\angle CBA=30^\circ\;,\;\;\angle CHA=90^\circ\;$$ 
Remembering that in a $\;30-60-90\;$ triangle the length of the leg in front of the angle of $\;30^\circ\;$ equals half the length of the hypotenuse, we get:
$$MH=\frac12CM=\frac12BM=\frac14AB\implies\frac{AB}{CM}=2\implies$$
$$\frac{S_{\Delta ABC}}{S_{\Delta MCH}}=\left(\frac{AB}{CM}\right)^2=\implies S_{\Delta ABC}=4\cdot S_{\Delta MCH}=4K$$
A: $BM=AM=CM$ property of the median $CM$ of the hyphotenuse in a right triangle $\triangle ACB$ with $\angle C=90^{\circ}$ is called "magnificent triple". In this question we also have, $\angle MCA=60^{\circ}.$ The worst part of the question is to see that $\triangle ACM$ is equilateral! (Hint: Drop a perpendicular from $M$ to $CA$). So, $(MCH)=(ACH)=\frac{1}{2}(ACM)=\frac14(ABC)\implies (ABC)=4K$.
A: 
A rough sketch of school trigonometry  $30^0,60^0,90^0$ triangle. Right angle at D is trisected.
$\Delta DOB$ is equilateral. Semicircle arc $ADB$ of unit radius. Line lengths
$$\{f.k,h,g\}= \{2,1,\dfrac{\sqrt{3}}{2},1 \}.$$
Small triangle area
$$= \frac{\sqrt 3}{2}\cdot \frac12 \cdot \frac12=\frac{\sqrt{3}}{8} $$
Large triangle area
$$=  \frac12 \cdot 1 \cdot {\sqrt 3} =\frac{\sqrt{3}}{2} $$
which is four times the smaller one.
