The problem is from my textbook, the topic is "A midline of a triangle":
Given a triangle $ABC$ and points $D$ and $E$ such that $∠ADB =∠BEC = 90°$. Prove that $DE ≤ \frac{1}{2}P \triangle ABC$.
My solution:
As it's known, a median of a right triangle, lowered from a right angle, is equal to a half of a hypotenuse, so let's lower two medians:
$DK$ onto $AB$ => $DK=AK=KB$
and
$EM$ onto $BC$ => $EM=BM=MC$.
$KM$ itself is a midline of $\triangle ABC$, so:
$\frac{1}{2}P \triangle ABC$ = $\frac{1}{2}AC + AK + KC$ = $DK + KM + ME$
What I'm left with is a polygonal chain of the $DK$, $KM$ and $ME$ line segments and I'm not sure how to finish the proof. "Each side of a triangle is smaller than the sum of its two other sides.", as my textbook states, but I'm not sure how to apply it here.
Any suggestions ?
Instead of attacking the question posed in OP’s problem statement, let us consider $\mathrm{Fig.\space 1\it{a}}\space$ given above. There, we show that segment $UV$ is shorter than or equal the length of the segment $GH$.
$\mathrm{Fig.\space 1\it{a}}\space$ shows two circles intersecting at $S$ and $T$. The red circle has its center at $P$ while that of the green circle is at $Q$. Line drawn through $P$ and $Q$ cut the two circles at $G$ and $H$. The point $W$ lies on $GH$ and a line through $W$ cut the circles at $U$ and $V$.
We apply the triangle inequality to the triangle $UPW$ to obtain, $$WU \le PU+PW=WG.\tag{1}$$
Similarly, we can prove that, $$WV\le WH.\tag{2}$$
We can use (1) and (2) to show that, $$WU+WV\le WG+WH \qquad\rightarrow\qquad UV\le GH.$$
Now, we apply this to the configuration depicted in $\mathrm{Fig.\space 1\it{b}}\space$ to show that the inequality mentioned in your problem statement holds.
$\mathrm{Fig.\space 1\it{b}}\space$ is an extended version of OP’s sketch, in which we have added two circles with centers at the midpoints $K$ and $M$ of the sides $BA$ and $BC$ respectively. Besides, the line segment $GH$, which passes through $K$ and $M$, is also included and this line is going to play a vital role in our proof of OP’s inequality. Let’s determine the length of $GH$ as described below. For brevity we denote the semiperimeter of $\triangle ABC$ by $s$. $$GH=GK+KM+MH=BK+\dfrac{AC}{2}+BM=\dfrac{BA}{2}+\dfrac{AC}{2}+\dfrac{BC}{2}=s$$
According to the lemma, which we have already proven, we have, $$DE\le GH=s.$$
The equality holds when $D$ and $E$ coincide with $G$ and $H$ respectively.
The story does not end here. In order to prove that the mentioned inequality is valid in general, we need to prove another version of the lemma (see $\mathrm{Fig.\space 2\it{a}}$) and apply it to the corresponding scenario of the inequality (see $\mathrm{Fig.\space 2\it{b}}$) to show that it holds.