$ABC$ is an isosceles triangle with base $AB$. Suppose that $AB = 5$ cm and the height of the triangle is $10$ cm. A square is placed inside the triangle $ABC$ so that one of its vertices is the midpoint of $AB$. Find the maximum area of the square in cm$^2$.
Can somebody help me with this question? I have been trying to solve this but I couldn't find any way to do so. Thank you.
Here is an approach.
First, try to find the largest square inside by experiments and observation, drawing the triangle and various squares in it.
Suppose $Q$ is the center of $AB$. Let the square be $PQRS$,
It looks that if the square is symmetric to line $CQ$ with $P$ on $CA$ and $R$ on $CB$, the square is as large as possible. Let $O$ be the center of the square. Then $C, O, Q$ are collinear. $OP\parallel QA$. We have $$\frac{OP}{QA}=\frac{CO}{CQ}.$$ Let $OP=x$. $$\frac{x}{\frac52}=\frac{10-x}{10}.$$ $$x=2.$$ Hence the area of $PQRS$ is $2x^2=8$.
Now our task is to prove $PQRS$ as described above does have the largest area among all possible squares.
Since $QP=2\sqrt2$, let us prove that any right isosceles inside the triangle with its right angle at $Q$ (which is a half of the square) must have its two equal sides no longer than $2\sqrt2$. Suppose the right isosceles is $\triangle XQY$. WLOG, suppose $\angle XQA\le 45^o$. Now we can prove that $$QX\le2\sqrt2.$$