Let $ ABCD$ be a convex quadrilateral such that $ AB$ and $ CD$ are not parallel and $ AB=CD$. The midpoints of the diagonals $ AC$ and $ BD$ are $ E$ and $ F$, respectively. The line $ EF$ meets segments $ AB$ and $ CD$ at $ G$ and $ H$, respectively. Show that $ \angle AGH = \angle DHG$.
Problem
Source: MEMO 2009, problem 3, single competition
Tags: geometry, rhombus, geometric transformation, homothety, geometry proposed
livetolove212
01.10.2009 12:47
Let $ I=AB\cap CD, J,K$ be the midpoints of $ AD, BC$. A well-known result: $ JK$ is parallel to the bisector $ l$ of angle $ GIH$. But $ JFKE$ is a rhombus then $ JK\perp EF\Rightarrow l\perp GH$. Thus $ IGH$ is the isosceles triangle.
plane geometry
01.10.2009 12:55
It is easy to see PFQE is an rhombus PQ⊥EF => PQ is the bisector of ∠FPE Denote IJ is the bisector of ∠BJCS => IJ∥PQ => IJ⊥EF => JGH is an isosceles triangle
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kasym
08.12.2009 11:43
I used homotety of triangles FPE and GJH
steppewolf
14.08.2015 16:12
Let $S$ be the point of intersection of lines $AB$ and $CD$. Using Menelaus theorem on the line containing $E$,$F$,$G$ and $H$ with respect to $\triangle ASC$ we get $\frac{HS}{CH}\frac{CE}{AE}\frac{AG}{GS} = 1$. Using Menelaus again this time on $\triangle BSD$ gives $\frac{HS}{DH}\frac{DF}{BF}\frac{BG}{GS} = 1$. Now $CE = AE$ and $DF = BF$, so $\frac{HS}{GS}\frac{AG}{CH} = \frac{HS}{GS}\frac{BG}{DH}$. Hence $\frac{AG}{BG} = \frac{CH}{DH}$, but $AB = CD$, which means $AG = CH$ and $BG = DH$. Now because $\frac{HS}{GS}\frac{AG}{CH} = 1$ we get that $HS = GS$, so the angles $\angle AGH$ and $\angle DHG$ are indeed equal.