Angle chasing problem | Mathematics | Page 1

humble pelicanBOT Oct 21, 2023, 3:27 PM

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shy ferry Oct 21, 2023, 4:17 PM

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Maybe an idea (need to try it)

Consider a parallel to (AB) that passes through G. Call it (d).

Consider the translation of vector $\vec{CB}$ that sends C on B.

Lines (BH) and (CG) are parallel:
Angle BHI = 117
And if you extend line CG, it intersects (HI) at K.
Since angle CGK = 180
Then angle KGI = 180 - 162 = 18
And GKI = 180-18-45 = 117

So lines (CG) and (BH) are parallel

Therefore the translation of vector $\vec{CB}$ sends G on G'.

The quadrilatetal GG'BC is a parallelogram

Angle CBG is sent on BB'G' with B' the image of B by the translation.

Now I have an intuition that maybe G' is the point we see when the right circle intersects BDH.

To be continued

gloomy brookBOT Oct 21, 2023, 4:17 PM

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Daddy_314

shy ferry Oct 21, 2023, 4:54 PM

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Thinking out loud:
(K'F) is perpendicular to (CF)

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Because CFK' is inscribed

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In a circle of diameter one of its sides

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Sorry I didnt mean K

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I meant the extension of (CK)

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This gives you a right triangle above

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If you call K' the intersection of GK and the circle
K'F perpendicular to CF

#Angle chasing problem