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!

JovenlyCosmo Nov 22, 2023

#1**0 **

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,

51422598=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 $BI = 7*sqrt(2)/5$.

bingboy Nov 22, 2023