Equiangular Right Triangles are Similar/Proof 2

Theorem

Equiangular right triangles are similar.

Proof

Let $ABC$ be an arbitrary right triangle with $\angle ABC$ a right angle.

Construct a straight line from $A$ parallel to $BC$.

In the words of Euclid:

Through a given point to draw a straight line parallel to a given straight line.

(The Elements: Book $\text{I}$: Proposition $31$)

Construct a second straight line from $C$ parallel to $AB$, meeting the first straight line at $D$.

By Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel:

$\Box ABCD$ is a parallelogram.

By definition of rectangle, and from Rectangle is Parallelogram:

$\Box ABCD$ is a rectangle.

By construction, $AC$ is a diameter of $\Box ABCD$.

Let $H$ be an arbitrary point on $AC$.

Constuct $EHF$ parallel to $BC$ intersecting $AB$ at $F$ and $CD$ at $H$.

Constuct $DGI$ parallel to $AB$ intersecting $AD$ at $G$ and $BC$ at $I$.

Hence by construction $EHF$ and $DGI$ are both parallel to the sides of $\Box ABCD$.

By Triangle Side-Side-Side Congruence:

$\triangle ABC \cong \triangle ADC$

so:

$\angle BAC = \angle DCA$

$\angle DAC = \angle ACB$

By Parallelism implies Equal Corresponding Angles:

$\angle AHF = \angle ACB$

Because the diameter of each is shared, by Triangle Angle-Side-Angle Congruence:

$\triangle AFH \cong \triangle AGH$

$\triangle HIC \cong \triangle HEC$

For the same reasons and also by construction:

$\triangle ABC \cong \triangle ADC$

By definition of congruence, $\Box ABCD$ is composed of two parts of equal area:

$\map \AA {\triangle AFH} + \map \AA {\Box FBIH} + \map \AA {\triangle HIC}$

$\map \AA {\triangle AGH} + \map \AA {\Box GHED} + \map \AA {\triangle HEC}$

By subtraction, we obtain:

$\map \AA {\Box FBIH} = \map \AA {\Box GHED}$

From Area of Rectangle:

$FH \cdot HI = GH \cdot HE$

Rearranging:

$\dfrac {FH} {GH} = \dfrac {HE} {HI}$

Thus, corresponding sides within each right triangle have equal ratios.

and also corresponding sides compared between two right triangle have equal ratios:

$\dfrac {FH} {HE} = \dfrac {GH} {HI}$

We can extend this result to the hypotenuse:

In the above diagram, the altitude of $\triangle ABC$ is $CD$.

By Perpendicular in Right-Angled Triangle makes two Similar Triangles:

$\triangle ADC \sim \triangle CDB$

By what we have shown above:

$\dfrac {AC} {BC} = \dfrac {AD} {CD} = \dfrac {CD} {BD}$

Rearranging:

$\dfrac {BC} {CD} = \dfrac {AC} {AD}$

which yields the result.

$\blacksquare$