Let $a$ and $b$ be positive real numbers. Let $C$ be a point on the positive $x$–axis and $D$ a point on the positive $y$-axis so that $CD = a + b$. The point $P$ on the line segment $CD$ satisfies $PD = a$ and $PC = b$. Prove that $P$ is on the ellipse given by the equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
Hint given is to consider $2$ similar right angle triangles.
I presume that is reference to the $2$ triangles in the diagram but am not sure. All help appreciated.
Let $O$, the center of the ellipse, be the origin of coordinates. Let $H$ be the projection of $P$ on the $x$ axis, and $K$ the projection of $P$ on the $y$ axis. Let $(x,y)$ be the coordinates of $P$, so that $H$ has coordinates $(x,0)$ and $K$ has coordinates $(0,y)$. Let the coordinates of $C$ be $(X,0)$ and of $D$ be $(0,Y)$. Then $X^2+Y^2=CD^2=(a+b)^2$.
Use Thales' theorem in triangles $PHC$ and $DOC$:
$$\frac{X-x}X=\frac{b}{a+b}$$
$$x=X-\frac{bX}{a+b}=\frac{aX}{a+b}$$
Likewise in triangles $DKP$ and $DOC$:
$$\frac{Y-y}{Y}=\frac{a}{a+b}$$ $$y=Y-\frac{aY}{a+b}=\frac{bY}{a+b}$$
Now,
$$\frac{x^2}{a^2}+\frac{y^2}{b^2}=\frac{X^2+Y^2}{(a+b)^2}=1$$
And the point $P$ lies on the ellipse of equation $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$.
This property gives a very simple method to draw an ellipse: the paper strip method.