Geometric way to construct image of the image of a point bu given homothety

Question

Let $ABC$ be a right triangle at $A$ and $G$ a point interior to it.

$D$ is the where lines $(AG)$ and $(BC)$ meet.

Let $h$ be the homothety of center $A$, that transforms $G$ to $D$.

I wonder whether there is a geometric way to place $D'$ the image of $D$ by $h$.

Thanks.

Answer 1

Just similar triangles your way to get the ratio $ \frac{AG}{AD}$ propagated.

As an example:

• Draw another line $\ell$ through A.
• Pick any point $E$ on $\ell$.
• Connect $ED$.
• Draw line parallel to $ED$ passing through $G$, intersecting $\ell$ at $F$.
• Draw $FD$
• Draw line parallel to $FD$ passing through $E$, intersecting $AGD$ at $H$.
• Since $\frac{AD}{AH} = \frac{AF}{AE} = \frac{AG}{AD}$, $H$ is the image of $D$.

Answer 2

With compass center at $A$, make circular arc through $G$ down to segment $\overline{AC}$, to produce point $\hat{G}$ so that $|AG| = |A\hat{G}|$. Analogously, produce point $\hat{D}$ on segment $\overline{AC}$ so that $|AD| = |A\hat{D}|$.

Now draw in $\hat{G}D$ and construct parallel through $\hat{D}$. This line will intersect line $\overline{AD}$ at the point $D'$ that you desire. Why? Similar triangles $\triangle A\hat{G}D \sim \triangle A\hat{D}D'.$