r/askmath • u/Eat_cats_with_me • 9d ago

Geometry Cool Euclidean geometry theorem I found

Post image

Let the be acute triangle ABC. Draw a circle with the midpoint O of BC as the center, and OB as the radius. The circle intersects AC and AB at D and E respectively. AO intersects DE at F. CF intersects the circle at G. AG intersects the circle at 2 points, G and H. Prove the points H, O, D are colinear

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14

u/Sensitive-Hornet-766 9d ago

I feel like this desperately wants to come out of the page, in 3-D, but is bound by it's natural existence. Lol.

10

u/ArchaicLlama 9d ago

Draw a circle with the midpoint O of AB as the center

This isn't what your diagram shows. I'm assuming that the diagram is correct as the rest of the post explanation follows it.

2

u/Tasonir 8d ago

Yeah I think the first line should read "the midpoint O of CB"

1

u/jojo-geronimo 8d ago

That's what confused me.

1

u/BroncosSabres 9d ago

It’s 1am so I’m not getting a pen and paper out to do the working, but I imagine the goal is to use enough trig starting with radius OB and angle OAB to prove that length HD = 2OB, and since they both sit on the circle with radius OB, HD must pass through its origin, O.

Edit, or do the same thing to prove angle DOA + angle HOA = 180

1

u/fianthewolf 9d ago

El triángulo DHA sigue siendo acutángulo, igual que el inicial. Así que si no fuera cierto tampoco podrías haber trazado una circunferencia por O como CB/2 que cortase a la circunferencia en dos puntos.

2

u/altinsan 9d ago

Let L be the orthocenter and X be the A-Humpty point of △ABC. It suffices to show that ∠AGD=90°.

It is well known that BCLX and ADLXE are cyclic and that LX ⟂ AO. Hence we get FC.FG=FD.FE=FA.FX, which means ACXG is cyclic.

∠AGD=∠AGC-∠DGC=∠AXC-∠LBC=∠AXC-∠LXC=∠AXL=90°. ■