How to prove that $DE=EF +DG$ from this following triangle problem?

Question

Given a right triangle $ABC$, where $C$ is a right angle.

We choose points $G$ at $AC$ and $F$ at $BC$, and $D$ and $E$ at $AB$.

We draw right triangles $AGD$ and $EBF$, such that $\angle AGD= \angle BFE= 90^{\circ}$, and $AE=AC$ and $BD=BC$.

Prove that $DE=EF +DG$?

I'm clueless about how to start from. I think we need to use similarity, but I still don't get it. Please explain your answer, and insert your picture to make me more understand. Thanks

Answer 1

Note that the figure is completely determined by the requirements. We do not get to "choose" points. Here's a re-statement of the problem:

Given right triangle $\triangle ABC$, with right angle at $C$, locate points $D$ and $E$ on $\overline{AB}$ such that $\overline{BD}\cong \overline{BC}$ and $\overline{AE}\cong\overline{AC}$. Let $F$ and $G$ be the feet of perpendiculars from $E$ to $\overline{BC}$ and from $D$ to $\overline{AC}$.

Define segment lengths as follows: $$\begin{align} a := |\overline{BC}| \qquad b := |\overline{CA}| \qquad c &:= |\overline{AB}| \\ p := |\overline{DG}| \qquad q := |\overline{EF}| \qquad r &:= |\overline{DE}| = |\overline{AE}|-|\overline{AD}| = b - (c-a) = a + b - c \end{align}$$

Now, $$\triangle ADG \sim \triangle ABC \quad\implies\quad \frac{p}{c-a} = \frac{a}{c}$$ $$\triangle EBF \sim \triangle ABC \quad\implies\quad \frac{q}{c-b} = \frac{b}{c}$$ Therefore, $$p + q = \frac{a(c-a)}{c} + \frac{b(c-b)}{c} = \frac{ac + bc - (a^2+b^2)}{c} = \frac{ac + bc - c^2}{c} = a + b - c = r$$