Equiangular Triangles are Similar/Proof 2

Theorem

Let $\triangle ABC$ be any triangle.

Let $\triangle ADE$ be equiangular with it.

Draw the triangles so they coincide at one vertex and the bases are parallel.

$BC \parallel DE$

Draw the altitude $AFG$ to both $\triangle ABC$ and $\triangle ADE$.

(For an obtuse triangle, the altitude need not lie inside $\triangle ADE$).

$\triangle ABF$, $\triangle ADG$, $\triangle ACF$ and $\triangle AEF$ are all right triangles.

By what we had above: $\triangle ABF$ and $\triangle ADG$ are equiangular.

The same is true of $\triangle ACF$ and $\triangle AEG$.

$\dfrac {AB} {AD} = \dfrac {AF} {AG}$

$\dfrac {AC} {AE} = \dfrac {AF} {AG}$

Rearranging:

$\dfrac {AB} {AD} = \dfrac {AC} {AE}$

We have equal ratios of sides.

But we can do this for any two sides of the two triangles.

$\blacksquare$