If $\triangle ABC$ is isosceles and right-angle, with $AB=AC=x$ and $M,N$ are the midpoints of $AC$ and $AB$ respectively, find the area of $\triangle MON$
My solution goes as follows, If $D$ is a point on $CB$ such that $MD\perp CB$, we have that the area of $\triangle CMB$ is $\frac{x^2}{4}$. Hence $\frac{MD*x\sqrt{2}}{2}=\frac{x^2}{4}$ we then from this find the height of $\triangle COB$ and $\triangle AMN$, and since we also have the height of $\triangle CAB$ from $CB$, so from this we get the height of $\triangle MON$ and hence we have that its area is $\frac{x^2}{24}$. My solution is very long and tedious, is there a faster and easier approach to this problem?
Best Answer
Note that the triangles $OMN$ and $OBC$ are similar with $\frac{ON}{OC}=\frac12$. Then $$[OMN]= \frac{ON^2}{OC^2}[OBC]=\frac14\cdot \frac23 [NBC]= \frac16\cdot\frac12[ABC]=\frac1{24}x^2 $$