# Geometry: Proof related to right angled isosceles triangle

ABC is a right angled isosceles triangle. If AD is a bisector of angle BAC then prove that AC + CD = AB.

The right angled isosceles triangle

The right angle is at C.

• Is the right angle located in $C$? – Dr. Sonnhard Graubner Oct 6 '18 at 12:51
• What are your own thoughts on this problem? – Matt Oct 6 '18 at 12:53
• Are you sure of that? – Dr. Sonnhard Graubner Oct 6 '18 at 12:55
• Yes I am sure. The right angle is at C. @Dr. Sonnhard Graunber – Gaurav Mishra Oct 6 '18 at 12:57
• I figured out each individual angle but other than that I don't have anything. @Matt – Gaurav Mishra Oct 6 '18 at 12:58

## 2 Answers

After a Theorem of the angle bisector we get $$BD:DC=\sqrt{2}a:a$$ or $$\frac{a-DC}{DC}=\sqrt{2}$$ simplifying this we get $$DC=\frac{a}{\sqrt{2}+1}$$ so $$a+\frac{a}{\sqrt{2}+1}=a$$ and this is true since $$\sqrt{2}a+2a=2a+\sqrt{2}a$$ Where $$AC=BC=a$$ and $$AB=\sqrt{2}a$$

Since this is an isosceles right triangle the two acute angles are $$\pi/4$$ radians. We can, without loss of generality, take sides AC and BC to have length 1 so that AB has length $$\sqrt{2}$$. Bisecting angle A gives an angle of $$\pi/8$$ radians and DC has length $$\tan(\pi/8)$$ $$= \frac{sin(\pi/4)}{1+ cos(\pi/4)}= \frac{\frac{\sqrt{2}}{2}}{1+ \frac{\sqrt{2}}{2}}= \frac{\sqrt{2}}{2+ \sqrt{2}}$$.

$$AC+ CD= 1+ \frac{\sqrt{2}}{2+ \sqrt{2}}= \frac{2 + 2\sqrt{2}}{2+ \sqrt{2}}= \frac{4+ 4\sqrt{2}- 2\sqrt{2}- 4}{4- 2}= \frac{2\sqrt{2}}{2}= \sqrt{2}$$