# Geometry Problem 2

$$ABCD$$ be a convex quadrilateral. Let $$E = AB \cap CD$$, $$F = AD \cap BC$$, $$P = AC \cap BD$$, and let $$O$$ the foot of the perpendicular from $$P$$ to the line $$EF$$. Prove that $$\angle BOC = \angle AOD.$$

Note by Alan Yan
2 years, 7 months ago

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Have u framed this question ? Because this question is somehow similar to the 2003 INMO P1.

- 2 years, 7 months ago

This is a China TST

- 2 years, 7 months ago

Sorry friend but i do not consider my level to be that of the China TST. Even India TST is far away. So could u pls post some solvable problems ? Some NMO level(solvable)

- 2 years, 7 months ago

Okay. I will post another. But in the meantime, here is my solution.

It suffices to prove $$\angle COP = \angle AOP$$ and $$\angle BOP = \angle DOP$$. Extend $$BD$$ to intersect line $$EF$$ at $$S$$. Extend $$AC$$ to intersect $$EF$$ at $$T$$. We have that $$(BPDS)$$ and $$(ETFS)$$ are harmonic bundles. $$(BPDS$$ harmonic and $$\angle POS = 90^{\circ}$$ proves that $$\angle BOP = \angle POD$$.

$$E(BPDS)$$ harmonic implies $$E(APCT)$$ harmonic. This information along with $$\angle POT = 90^{\circ}$$ proves $$\angle COP = \angle AOP$$. Hence, we are done.

- 2 years, 6 months ago