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.
$\blacksquare$