Olympiad Question based on Centres of triangle

[night robin]

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[grizzled sapphire]

consider areas

if you can prove that all 6 small triangles have equal area then you are very close to done

[grizzled sapphire]

have you proven it?

[grizzled sapphire]

consider the areas of BCG and BCA now

they share the same base, so the height will account for the ratio

[grizzled sapphire]

and then that also happens to be the ratio between the lengths DG and DA, which isn't terribly hard to prove

well, BCG is one third the area of BCA

since one contains two small triangles and the other 6

[grizzled sapphire]

so the heights are also in the ratio 1:3

and this means DA = 3DG

so GA is twice DG

[grizzled sapphire]

nice

[grizzled sapphire]

does this answer your question?

hm

does it have to be proven using apollonius?

[grizzled sapphire]

BCEF cannot be cyclic unless AB = AC

it may also be worth noting that 10 = DG = DB = DC and as a result BC = 20

i believe cosine law finishes

[grizzled sapphire]

(this is a direct consequence of <BGC being a right angle)

i thought you were trying to prove the 2:1 medians ratio so i was off on the wrong track altogether

sorry about that

[plucky cairn]

@brazen trail has given 1 rep to @foggy socket

[grizzled sapphire]

thank you