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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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. – Michael Mendrin · 2 years, 11 months ago

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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. \( \angle AGB = \angle BDO = 90 ^ \circ \) – Roger Lu · 2 years, 10 months ago

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– Mathh Mathh · 2 years, 10 months ago

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

Construct a model diagram with AC and BC AB as tangents – Führer Sy · 2 years, 10 months ago

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– Führer Sy · 2 years, 10 months ago

With any dimensions of radius and lengths of triangleLog in to reply

AP PG GB – Shanzkie Vargas · 2 years, 10 months ago

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90 – Führer Sy · 2 years, 10 months ago

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I took an equilateral triangle and using coordinate geometry found out the value of the angle. Is the method legitimate? – Kartik Raj · 2 years, 10 months ago

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– Mathh Mathh · 2 years, 10 months ago

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.Log in to reply

There's no line that connects with A and G. Therefore, there is no angle AGB. – Kent Arvin Aquino · 2 years, 11 months ago

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@Kent Arvin Aquino – Elliott Macneil · 2 years, 10 months ago

There doesn't need to be a line connecting two vertices for an angle to countLog in to reply

– Ashay Wakode · 2 years, 9 months ago

It's from INMO ,right.Log in to reply

– Ashay Wakode · 2 years, 9 months ago

To Yankees rariaLog in to reply