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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.

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  • $\begingroup$ Is the right angle located in $C$? $\endgroup$ – Dr. Sonnhard Graubner Oct 6 '18 at 12:51
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    $\begingroup$ What are your own thoughts on this problem? $\endgroup$ – Matt Oct 6 '18 at 12:53
  • $\begingroup$ Are you sure of that? $\endgroup$ – Dr. Sonnhard Graubner Oct 6 '18 at 12:55
  • $\begingroup$ Yes I am sure. The right angle is at C. @Dr. Sonnhard Graunber $\endgroup$ – Gaurav Mishra Oct 6 '18 at 12:57
  • $\begingroup$ I figured out each individual angle but other than that I don't have anything. @Matt $\endgroup$ – Gaurav Mishra Oct 6 '18 at 12:58
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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 $$AB=\sqrt{2}a$$

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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} $

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