In the given figure, P and Q are point of contact. O is the incentre. Line BO produced, meets PQ at G. Find the value of angle AGB?

The reason i shared the question as a note is that i don't know the answer. Try it out, & if you solve, please post a rigorous solution rather than mere guess.

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## Comments

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Baffling Geometry

With lots of angle-chasing, it can be worked out that, given angles \(a, b\), the following angles are

\(\angle OAP=90-a-b\)

\(\angle OGP=90+a+b\)

\(\angle AOG=90-a\)

\(\angle APG=90+a\)

Hence, \(APGO\) is a cyclic quadrilateral. Moreover, since \(APO\) is a right triangle, \(AO\) is the diameter of the circle that circumscribes the cyclic quadrilateral \(APGO\). Hence, \(AGO\) is also a right triangle, and we have our answer.

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Here is another way to prove it. It is easy to see \( \angle PQC = {\angle BAC} /2 + {\angle ABC}/2 \). So you can get \( \angle BGQ = {\angle BAC} /2 = \angle BAO\), therefore \( \triangle BGQ \sim \triangle BAO \) and \( BG \times BO = BQ \times BA \). Let the point of contact of incircle with AB is D, we have \( BG \times BO = BD \times BA \). So AGOD is cyclic. \( \angle AGB = \angle BDO = 90 ^ \circ \)

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You haven't defined the point \(D\) in this proof. But it is simply the point touching the circle on the segment \(AB\).

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AP PG GB

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Construct a model diagram with AC and BC AB as tangents

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With any dimensions of radius and lengths of triangle

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I took an equilateral triangle and using coordinate geometry found out the value of the angle. Is the method legitimate?

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You've only found it when the triangle is equilateral. You haven't proved that the angle is the same for all triangles. You can't assume that all the triangles have the same angle just because the question asks for a single unique numeric value.

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90

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There's no line that connects with A and G. Therefore, there is no angle AGB.

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There doesn't need to be a line connecting two vertices for an angle to count @Kent Arvin Aquino

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It's from INMO ,right.

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