Unique RHS Triangle

Theorem: Given the length of hypotenuse and any other side of a right triangle, a unique triangle can be formed.

prerequisites:
Isosceles triangle theorem (proof)
Exterior angle property of a triangle (proof)

Proof:

Let $\triangle ABC$ be a right triangle, right angled at B, with the given length of hypotenuse AC and a side BC. If possible, let $\triangle A'BC$ be another right triangle, right angled at B, such that A'C = AC. Since, $\triangle A'BC$ is also right angled at B, therefore A' lies on the line AB. Let us first assume that AB $<$ A'B

Now in $\triangle ACA'$,

$\qquad\quad AC = A'C\qquad\qquad\qquad\qquad\qquad\;\;\;\text{(given)}$

Hence $\triangle ACA'$ is an isosceles triangle. Thus, by the isosceles triangle theorem,

$\qquad\quad\angle CAA' = \angle CA'A\qquad\qquad\qquad\qquad\cdots\text{(1)}$

Now in $\triangle ABC$,

$\qquad\quad\angle B = 90^o\\ \qquad\quad\angle BCA = \theta\\ \therefore\;\quad\;\;\angle CAA' = 90^o+\theta\qquad\qquad\qquad\qquad\text{(By exterior angle property of a triangle)}$
$\Rightarrow\quad\;\;\angle CA'A = 90^o + \theta\qquad\qquad\qquad\qquad\text{(from $(1)$)}$

In $\triangle ACA'$, let $S$ be the sum of all the interior angles,

$\begin{align}\therefore\quad\;\; S &= \angle CAA'+\angle CA'A + \angle ACA'\\ &= 90^o+\theta+90^o+\theta+\angle ACA'\\ &= 180^o + 2\theta + \angle ACA'\\ &> 180^o\end{align}$

But this is a contradiction, as by angle sum property of a triangle, $S = 180^o$.
Hence, AB $\not<$ A'B.

By similar analysis, taking $\triangle A'BC$ instead of $\triangle ABC$, it can be shown that AB $\not>$ A'B.

Hence AB $=$ A'B. Therefore, $\triangle ABC$ is unique.

Q.E.D.

Recommended:
Unique SSS triangle
RHS congruence
Pythagoras Theorem