In this note, I have a selection of questions for you to prove. It is for all levels to try. The theme is geometry. Good luck.

1) Scalene triangle \(ABC\) is right-angled at \(A\). The tangent to the circumcircle of triangle \(ABC\) at point \(A\) intersects \(BC\) at \(X\). Let the points of contact of the incircle of triangle \(ABC\) with sides \(AB\) and \(AC\) be \(E\) and\(F\) respectively. Let \(EF\) and \(BC\) at \(Y\) and \(AX\) at \(Z\).

Prove that triangle \(XYZ\) is both obtuse and isosceles.

2) Let \(ABCD\) be a square with centre \(F\). Let \(DFCE\) be a square and let \(BEHG\) be the square containing \(C\) in its interior.

Prove that \(C\) is the midpoint of \(AH\).

3) In triangle \(ABC\) it is known that \(\angle BAC = 2 \angle ACB\) and \(2 \angle ABC = \angle BAC + \angle ACB\). The bisector of angle \(ACB\) intersects \(AB\) at \(E\). Let \(F\) be the midpoint of \(AE\). Let \(D\) be the foot of the perpendicular from \(A\) to \(BC\). The perpendicular bisector of \(DF\) intersects \(AC\) at \(G\).

Prove that \(AG = CG\).

4) Point \(D\) lies outside circle \(\Gamma\). A line \(\mathit{l}\) through \(D\) intersects \(\Gamma\) at points \(E\) and \(F\). Points \(A\) and \(B\) are the points of contact of the two tangents from \(D\) to \(\Gamma\). The line passing through \(B\) and parallel to \(\mathit{l}\) intersects \(\Gamma\) at \(G\).

Prove that \(GA\) intersects the segment \(EF\) at its midpoint.

5) Acute triangle \(ABC\) has circumcircle \(\Omega\). The tangent at \(A\) to \(\Omega\) intersects \(BC\) at \(D\). Let \(E\) be the midpoint of the segment \(AD\). Let \(F\) be the second intersection point of \(BE\) with \(\Omega\). Let \(G\) be the second intersection point of \(DF\) with \(\Omega\).

Prove that \(CG\) is parallel to \(DA\).

6) Prove that if 3 congruent circles pass through the same point, then their other three intersection points lie on a fourth circle with the same radius.

## Comments

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TopNewestHint:

1.) Angle chasing

2.) Pythagoras

6.) Consider the centroid of triangle formed by each center.

Too bad I'm not in mood of geometry right now. =..=" – Samuraiwarm Tsunayoshi · 2 years, 4 months ago

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– Sharky Kesa · 2 years, 4 months ago

Are you now?Log in to reply

– Samuraiwarm Tsunayoshi · 2 years, 4 months ago

Probably in a week or 2. btw Proofathon Geometry will be here in 2 days though. =="Log in to reply