olymmath

HMMT 2012 S6

  • Team 2

Let ABCA B C be a triangle with AB<ACA B<A C. Let the angle bisector of ∠A\angle A and the perpendicular bisector of BCB C intersect at DD. Then let EE and FF be points on ABA B and ACA C such that DED E and DFD F are perpendicular to ABA B and ACA C, respectively. Prove that BE=CFB E=C F.

Dataset text NuminaMath olympiads. Quoted with source.

Also HMMT 2012 T1.

Solutions

No solution is on olymmath yet. Official and human solutions come first when they arrive. AI solutions come after them, with a label.

Add a solution

Similar problems

These problems have the most words in common with this statement.

Locator hmmt/2012/s6 · Tag QMAASY2P68 · Edit this page on GitHub · Keys: j next, k previous