Source: 2018 Canadian Open Math Challenge Part C Problem 3 Consider a convex quadrilateral $ABCD$. Let rays $BA$ and $CD$ intersect at $E$, rays $DA$ and $CB$ intersect at $F$, and the diagonals $AC$ and $BD$ intersect at $G$. It is given that the triangles $DBF$ and $DBE$ have the same area. $\text{(a)}$ Prove that $EF$ and $BD$ are parallel. $\text{(b)}$ Prove that $G$ is the midpoint of $BD$. $\text{(c)}$ Given that the area of triangle $ABD$ is 4 and the area of triangle $CBD$ is 6, (C.)compute the area of triangle $EFG$.
Problem
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Tags: Comc, 2018 COMC
Piazolla13
06.12.2018 21:20
cms made a mistake because in part (c) if the area of triangle ABD is 4 it's impossible for the area of triangle CBD to be 6
MessingWithMath
07.12.2018 05:47
wait what? That isn't true. Answer for 3(c) is 120.
Immanuel Bonfils
11.12.2018 04:39
(a)Let's take the base $BD$ common to both triangles; The altitudes of $F$ and $E$ (distances to $BD$) must be the same, therefore $EF \| BD$. (b)If $JI\|BD$,where $J \in BF,\;I \in DE\;A\in JI$, similar triangles $BDF \sim JAF$ and $BDE \sim IAE$ gives $${{AJ}\over{BD}}={{AF}\over{FD}}\;,\;{{AI}\over{BD}}={{AE}\over{BE}}$$and $$BDA \sim AEF \implies\;{{AD}\over{AF}}={{BA}\over{AE}}\implies{{AE}\over{BE}}={{AF}\over{FD}}$$that with the previous expressions gives $AI=AJ$, giving the answer (b).
Immanuel Bonfils
11.12.2018 20:36
Is there enough information for part (c) ?
Immanuel Bonfils
13.12.2018 16:38
Could you, please, answer if something is missimg to part (c)?