Find The Angle

Geometry Level 3

In acute $\triangle ABC,$ $\angle BAC = 80^{\circ}.$ The internal angle bisectors of $\angle ABC$ and $\angle BCA$ intersect sides $CA$ and $AB$ at points $P,Q$ respectively. Let $I$ be the incenter of $\triangle ABC$ (the point where $BP$ and $CQ$ meet). Suppose the circumcenter of $\triangle PQI$ lies on the line joining $A$ and $I.$ Find $\angle ABI$ in degrees.

The answer is 25.

4 solutions

Jianzhi Wang
May 5, 2014

Firstly, draw line AI. Obviously AI must be the angle bisector of angle BAC. Since the centre of the circle lies on AI, then triangle AQI must be congrent to triangle API. (QI = PI, AP = AQ, Ai = AI [SSS]) So angle QAI = angle PAI and angle AIP = AIQ. Angle ABI = angle AIP - angle BAI = angle AIQ - angle CAI = angle ACI. This shows that angle ABC = angle ACB, which shows that triangle ABC is an isosceles triangle. Thus, angle ABI = 25 degrees.

Nice solution!

Sreejato Bhattacharya - 7 years, 1 month ago

Why does AP = AQ must hold true? ("Since the centre of the circle lies on AI". doesn't seem reason enough)

Daniel Garcia - 6 years, 10 months ago

Maharnab Mitra
May 5, 2014

I can be easily seen that the triangle satisfies the given conditions if it is an isosceles triangle. $\angle ABI$ , thus, becomes half of $\angle B$ .

Yes, but how do you know no other triangle works?

Sreejato Bhattacharya - 7 years, 1 month ago

Umm..... actually it seemed quite trivial to me. But in reality, I never tried to prove it.

Maharnab Mitra - 7 years, 1 month ago

point A , circum center of trianle PQI and I can come in line only when triangle ABC is isosceles with AB = AC

anurag spartan - 6 years, 11 months ago

Yes, but you need to prove that it must be an isosceles triangle and that only an isosceles triangle satisfy the condition.

Jianzhi Wang - 7 years, 1 month ago

Mohammed Tajuddin
May 10, 2015

the point 'i' is equidistant from B and C ( as the intersection of angle bisectors are equal from all three points) therefore ib =ic (=ai not necessary over here) angle b = angle c therefore angle 'a' + angle 'b' + angle 'c' = 180 hence angle b = 50 and angle abi is 25

Niranjan Khanderia
May 25, 2014

Let N be the circumcenter. Hence NA = NB = NC...............AI is also angle bisector.
Thus angles NBA = NAB = NAC = NCA =80/2 = 40. But NBC = NCS......Sum of all angles = 180. Hence NBC = 10. ....ABC = 40 + 10 = 50. IBA = ABC/2 = 50/2 = 25.

Find The Angle

The answer is 25.

4 solutions

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