# Angle bisector in a right angled triangle

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).

• "Other than"?? Why on earth would you think it didn't come from the sine or cosine of 45. Of course it comes from the sine/cosine of 45! Draw a picture and it'll be obvious. Apr 22, 2017 at 21:03

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

• Thank you for your answer. Did you find |CD| using the pythagorean theorem? Apr 23, 2017 at 5:19
• Yes, $\triangle CDE$ is right and isosceles, then $|CD|^2=x^2+x^2=2x^2$. Therefore we find $|CD|=\sqrt{2x^2}=x\sqrt 2$. Apr 23, 2017 at 9:12
• Diagonal of the square (CD) = √2 * side of the square Jun 15, 2020 at 9:33
• But how did you get from $\frac{b}{a} = \frac{x}{a - x}$ to $x = \frac{ab}{a + b}$? Dec 10, 2021 at 22:22
• @G. Kashtanov Hi, The part you ask is the solution of the 1st degree equation with one unknown. I hope that you know the cross product. Let's use this: $ax = b(a-x)$. Now we get $ax = ab - bx$ using the distributive property. Thus, $ax + bx = ab$ and $(a+b)x = ab$. Finally we find that $x = \dfrac{ab}{a+b}$. Feb 24, 2022 at 12:20

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 $$\bbox[5px,border:2px solid #C0A000]{d^2=\frac{2a^2b^2}{(a+b)^2}}\tag{5}$$

• Can you explain the ac/a+b and bc/a+b ? I dont find it on google
– Dini
Feb 8, 2020 at 6:53
• The Law of Sines says the two pieces of $c$ are of length $\frac{a}{\sin(D)}\sin(C/2)$ and $\frac{b}{\sin(\pi-D)}\sin(C/2)=\frac{b}{\sin(D)}\sin(C/2)$. The ratio of these lengths is $\frac ab$. That means the lengths are $c\frac{a}{a+b}$ and $c\frac{b}{a+b}$.
– robjohn
Feb 8, 2020 at 8:38

Alternative solution: Refer to the diagram below:

The Sine Theorem for $\Delta ACD:$

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

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

Just use the fact that area of a triangle PQR is PQsinx, where x is the included angle by P and Q . And here, sum of the areas of the two triangles (which are made by the angle bisector) is equal to 1/2*AB*BC(i.e. Area of ABC). sin45 will give 1/root2