In ABC find points X,Y,Z such that AXYZ is rhombus

Question

Question -

In ABC find points X,Y,Z on AB,BC,CA such that AXYZ is rhombus and area of AXYZ <= 1/2 AREA OF ABC

My try -

I know it very easy but I am not getting ...I take midpoints of sides and proved that opposite sides are equal by midpoint theorem but not able to prove adjacent sides are also equal ..

Answer 1

Construct bisector $AY$ of $\Delta ABC$.

Let $X\in AB$ such that $XY||AC$ and $Z\in AC$ such that $YZ||AB.$

Thus, $AXYZ$ is a rhombus.

Now, let $XY=k$.

Thus, since $\Delta XBY\sim\Delta ABC,$ in the standard notation we obtain: $$\frac{k}{b}=\frac{c-k}{c},$$ which gives $$k=\frac{bc}{b+c}.$$ Id est, we need to prove that: $$\left(\frac{bc}{b+c}\right)^2\sin\alpha\leq\frac{1}{4}bc\sin\alpha$$ or $$(b-c)^2\geq0,$$ which is obvious.