# Finding $\frac{DK}{DI}$

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.

• Attach the diagram to the question.
– user
Feb 14, 2021 at 21:41
• Done, sorry for neglecting that. Feb 14, 2021 at 21:49
• I think it would be useful to draw the line $AI$ as well.
– user
Feb 14, 2021 at 21:54
• 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? Feb 14, 2021 at 21:56
• Oh, right! Thanks, I got it! Feb 14, 2021 at 22:15

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: $$\color {red}{DI}=DK+AK-AI=4\sqrt{\frac{14}5}.$$
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 $$DK$$