+36 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!
+743 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 $BI = 7*sqrt(2)/5$.
