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Let \(I\) be the inradius of \(\triangle ABC\) where \(AB=25, AC=25, BC=14\). Find \(\overline{BI}\).

 

I got \(\frac{35}{4}\), can you guys check my answer?

 

Thanks!

 Nov 22, 2023
 #1
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Since AB=AC, △ABC is isosceles, so ∠CAB=∠CBA, and the two medians CD and BE intersect at the incenter I. Also, since CD is a median, D is the midpoint of BC, so BD=14/2=7.

Since △ABC is isosceles, CD is also an altitude of △ABC. Since CD is an altitude, ∠ICD=90∘. Since ∠ICD=90∘, rectangle ICBD is formed, so CI=BD=7. Since IC=7, CDBD​=21​.

Now, since CDBD​=21​, △ICD is similar to △ABC. Since △ICD is similar to △ABC,

ABCI​=BCCD​.

Substituting the given values, we have

257​=14CD​.

Solving for CD, we get CD=2514⋅7​=2598​.

Since CD=2AC​=2598​, AC=2⋅CD=2598​⋅2=25196​.

Since ∠CAB=∠CBA, △ABC is 45-45-90 triangle. Since △ABC is a 45-45-90 triangle, BC=2​⋅AC=2​⋅25196​=5249​​=5142​​.

Since BCCD​=21​,

5142​​2598​​=21​.

Simplifying both sides, we have

272​​=1425​.

Multiplying both sides by 2142​​, we get

72​=7175​.

Squaring both sides, we have

49⋅2=4930625​.

Solving for 2, we get 2=49625​.

 

Then $BI = 7*sqrt(2)/5$.

 Nov 22, 2023

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