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

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

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## Monday, July 6, 2015

### Geometry Problem 1128: Right Triangles, Medians, Parallel, Angles, Congruence, Half the measure

Labels:
angle,
congruence,
half,
median,
parallel,
right triangle

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http://s7.postimg.org/ccpzrcdez/pro_1128.png

ReplyDeleteObserve that quadrilateral DBEB1 is cyclic so α = α 1

Let B1M1 cut BE extension at N

Let b=∠ (MBC) => ∠ (MCB)= β

Since BM//B1M1 => ∠ (M1Nx)= β

And ∠ (CFE)= 180- β - α

And ∠ (M1B1C1)= β - α =∠ (M1C1B1)

In triangle GFC1 θ = 180-(180- β - α)-( β - α)= 2. α