1984 AMC 12 Problem 26

Attempt Problem 26 of the 1984 AMC 12 below, then check your answer against the professionally curated solution from LIVE by Po-Shen Loh. You can also try the full timed exam, view all 1984 AMC 12 solutions, or check the answer key.

All problems are used with official legal permission of the Mathematical Association of America (MAA).

26.

In the obtuse triangle ABC,ABC, AM=MB,AM=MB, MD⊥BC,MD\perp BC, EC⊥BC.EC\perp BC. If the area of △ABC\triangle ABC is 24,24, then the area of △BED\triangle BED is
99
1212
1515
1818
not uniquely determined
Answer: B
Concepts:triangle areamidpointarea decomposition
Small hint:
Draw segment MCMC and compare △DMC\triangle DMC with △DME\triangle DME
Big hint:
Those triangles have the same base MDMD and equal altitudes; also MM is the midpoint of ABAB
Solution:
Draw MC.MC. Since EC∥MD,EC\parallel MD, points EE and CC have equal perpendicular distances from MD,MD, so [DME]=[DMC].[DME]=[DMC]. Therefore [BED]=[BMD]+[DME]=[BMD]+[DMC]=[BMC]. \begin{aligned} [BED]&=[BMD]+[DME]\\ &=[BMD]+[DMC]\\ &=[BMC]. \end{aligned} Because MM is the midpoint of AB,AB, triangle BMCBMC has half the area of ABC.ABC. Thus [BED]=12.[BED]=12.
Therefore, the correct answer is B.
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