\(ABCD\) 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 \(\angle BOC = \angle AOD.\)

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)

@Shrihari B
–
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.

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

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This is a China TST

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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)

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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.

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