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

No vote yet

1 vote

×

Problem Loading...

Note Loading...

Set Loading...

## Comments

Sort by:

TopNewestHave u framed this question ? Because this question is somehow similar to the 2003 INMO P1. – Shrihari B · 1 year, 7 months ago

Log in to reply

– Alan Yan · 1 year, 7 months ago

This is a China TSTLog in to reply

– Shrihari B · 1 year, 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)Log in to reply

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. – Alan Yan · 1 year, 6 months ago

Log in to reply