In triangle ABC, it is known that AB = 12,BC = 13 and AC = 15. D is a point on AC such that the in-circle of triangle ABD and BCD have equal radii. Find AD:DC.
What i tried that as R is equal and R^2 = (S-A)(S-C)(S-B)/S so in this question
(let S1 be the semi-perimeter of triangle ABD and S2 of BCD)
(S1 - 12)(S1- AD)(S1-BD)/S1 = (S2-13)(S2-BD)(S2-DC)/S2
and then compare the terms. On comparing i get AD - DC = 1; and as AD + DC = 15 so
by this method i get AD:DC = 8:7 so is this method correct. Means can i compare LHS and RHS in this question the way i did. Or is there some other method to do this question.
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Smit Patel
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Solution
Lemma
Let $M$ be a point on the side $BC$ of $\triangle ABC$ such that the radiuses of the incircles of $\triangle ABM$ and $\triangle ACM$ are equal. Then $$AM=\frac{\sqrt{(b+c)^2-a^2}}{2},$$ where $a,b,c~~$ are the lengthes of $BC, CA, AB$ respectively.
As for the present problem, we obtain $$BD=\frac{\sqrt{(12+13)^2-15^2}}{2}=10.$$
Notice that $$\frac{S_{ABD}}{S_{BCD}}=\frac{AB+BD+AD}{BC+BD+DC}=\frac{AD}{DC}.$$
Namely, $$\frac{22+AD}{23+DC}=\frac{AD}{DC}.$$
But $$AD+DC=15.$$
Therefore, $$AD=\frac{22}{3},~~~DC=\frac{23}{3}.$$
It follows that $$\frac{AD}{DC}=\frac{22}{23}.$$
mengdie1982
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How did you found out that S (of ABD triange )/S(of BCD triangle)=(AB+BD+AD)/(BC+BD+DC)=AD/DC. – Smit Patel Aug 06 '18 at 16:49
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$S=pr$, where $p$ is the semi-perimeter. but $r$ is same.... – mengdie1982 Aug 06 '18 at 17:01
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What is S? And if S is area then how area of triangle ABD/BDC = AD/DC – Smit Patel Aug 06 '18 at 17:16
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don't you see the two triangle have the same altitude？ – mengdie1982 Aug 06 '18 at 17:34
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very sorry i got it – Smit Patel Aug 07 '18 at 07:55
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how did you prove that AM = square root of ((b+c)^2 -a^2)/2 – Smit Patel Aug 07 '18 at 07:58
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@SmitPatel See https://math.stackexchange.com/questions/2874815/show-that-am2-pp-a – mengdie1982 Aug 07 '18 at 10:50
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Thank you so much – Smit Patel Aug 08 '18 at 16:42