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In triangle $ABC,$ where $AB = 8, AC = 7,$ and $BC = 10,$ $I$ is the incenter. If $AI$ intersects $BC$ at $K$ and the circumcircle of $\triangle ABC$ at $D,$ find $\frac{DK}{DI}.$


I first drew a diagram, but I was unsure where to go from here.

enter image description here

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  • $\begingroup$ Attach the diagram to the question. $\endgroup$
    – user
    Feb 14, 2021 at 21:41
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    $\begingroup$ Done, sorry for neglecting that. $\endgroup$ Feb 14, 2021 at 21:49
  • $\begingroup$ I think it would be useful to draw the line $AI$ as well. $\endgroup$
    – user
    Feb 14, 2021 at 21:54
  • $\begingroup$ So far, I've found that $IK = \frac{3\sqrt{55}}{5}$ using Heron's Formula and the formula for an inradius. Is there a good way to find $DK$ though? $\endgroup$ Feb 14, 2021 at 21:56
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    $\begingroup$ Oh, right! Thanks, I got it! $\endgroup$ Feb 14, 2021 at 22:15

2 Answers 2

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I have doubts that you correctly compute the result, since the value $IK$ given in comments is incorrect. The correct result is: $$\frac{DK}{DI}=\frac23. $$ The details are given below.

Let $x,y,z$ being the distances from the vertices $A,B,C$ to the tangent points of the incircle. From the equations $x+y=c, y+z=a, z+x=b$ one obtains $x=\frac{b+c-a}2=s-a$, where $s$ is the semiperimeter. Then: $$AI=\sqrt{x^2+r^2}=\sqrt{(s-a)^2+\frac{(s-a)(s-b)(s-c)}s}=\sqrt {\frac {bc (s-a)}{s}}=2\sqrt{\frac{14}5}.$$ The angle bisector length is: $$AK=\sqrt{bc\left[1-\left(\frac a{b+c}\right)^2\right]}=\frac {\sqrt{4bc (s-a)s}}{b+c}=\frac{10}3\sqrt{\frac{14}5}.$$ By the power of point $K$ and angle bisector theorem:$$\color {red}{DK}=\frac{BK\cdot KC}{AK}=\frac{a^2\frac{bc}{(b+c)^2}}{AK}=\frac{a^2}{b+c} \sqrt {\frac{bc}{4 (s-a)s}}=\frac83\sqrt{\frac{14}5},$$ so that: $$\color {red}{DI}=DK+AK-AI=4\sqrt{\frac{14}5}.$$

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Observe that, $ID=DB=DC$.

Now, notice that, $\triangle AKB\sim \triangle ACD$ and hence $\frac{AD}{CD}=\frac{AB}{KB}$. Putting in $ID=CD$ will give the value of $ID$ and thereafter $DK$

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