Angle bisector in a right angled triangle

Question

In a right angled triangle, the legs adjacent to the right angle are equal to $a$ and $b$. Prove that the length of the bisector (of the right angle) is equal to $$\frac{a\cdot b\cdot \sqrt{2}}{a+b}.$$

While approaching this question, I was very puzzled as to how I would end up with this expression.

Additionally, I couldn't figure out where the $\sqrt{2}$ would come from, other than the sine or cosine of $45$ degrees (from the bisector).

Answer 1

An elemenraty solution: In the following figure $|BC|=a,|CA|=b,|AB|=c$ and $[CD]$ angle bisector. Let's draw square $CEDF$ and $|CE|=x$. So, $|BE|=a-x$ and $|CD|=x\sqrt2$. Now $\triangle ABC \sim \triangle DBE$ and $$\dfrac{b}{a}=\dfrac{x}{a-x} $$

Therefore $x=\dfrac{ab}{a+b}$ and $|CD|=\dfrac{ab\sqrt2}{a+b}.$

Answer 2

Alternative solution: Refer to the diagram below:

The Sine Theorem for $\Delta ACD:$

$$\frac{l_c}{\sin{CAD}}=\frac{b}{\sin{CDA}} \Rightarrow l_c=\frac{b \sin{CAD}}{\sin{(180^\circ}-(CAD+45^\circ))}=\frac{b\cdot\frac{a}{c}}{\sin{(CAD+45^\circ)}}=$$

$$\frac{ab}{c(\sin{CAD} \cdot \cos{45^\circ}+\cos{CAD}\cdot\sin{45^\circ})}=\frac{ab}{c(\frac{a}{c} \cdot \frac{1}{\sqrt{2}}+\frac{b}{c}\cdot\frac{1}{\sqrt{2}})}=\frac{\sqrt{2}ab}{a+b}.$$

Answer 3

In a General Triangle

Using the Law of Sines, we get that side $c$ is broken up as follows by the angle bisector:

Using the Law of Cosines we get $$ \left(\frac{ac}{a+b}\right)^2+d^2-\frac{2acd}{a+b}\cos(D)=a^2\tag{1} $$ and $$ \left(\frac{bc}{a+b}\right)^2+d^2+\frac{2bcd}{a+b}\cos(D)=b^2\tag{2} $$ Multiply $(1)$ by $b$ and $(2)$ by $a$ and add to get $$ (a+b)ab\left(\frac{c}{a+b}\right)^2+(a+b)d^2=(a+b)ab\tag{3} $$ Solving $(3)$ for $d^2$ yields $$ \bbox[5px,border:2px solid #C0A000]{d^2=ab\frac{(a+b)^2-c^2}{(a+b)^2}}\tag{4} $$

---

In a Right Triangle

Since, in a right triangle, $a^2+b^2=c^2$, $(4)$ becomes $$ \bbox[5px,border:2px solid #C0A000]{d^2=\frac{2a^2b^2}{(a+b)^2}}\tag{5} $$