Geometry Problem. Post your solution in the comments box below.

Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to see the complete problem 880.

## Thursday, May 23, 2013

### Problem 880: Triangle, Midpoints, Sides, Perpendiculars, Hexagon, Area

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Let S(PQR) be the area of ΔPQR.

ReplyDeleteForm ΔDEF.

Since ΔBFD, ΔFAE and ΔDEC are congruent,

and H, G, M are their orthocenters.

Thus

S(HFD)+S(FGE)+S(DEM) = S(DEF)

So

Area of hexagon = 2×S(DEF) = 12

Connect EF, FD and DE

ReplyDeleteNote that area(BFD)=Area(AFE)=Area(CED)=Area(EFD)=6

Triangles BFD, DEC and FAE are congruence (case SSS)

Triangle EDM congruence to triangle FBH ( case ASA)

Triangle FGE congruence to triangle BHD ( case ASA)

So Area(EGFHDM)= Area(EFD)+Area(FBD)= 12