## Monday, July 6, 2015

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

Geometry Problem. Post your solution in the comments box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the diagram below to enlarge it.

#### 5 comments:

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

Observe 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

1. arent alpha and beta are same angles?

2. No , Alpha =∠(ADA1)
beta=∠(CBA)

2. Let BE extended meet M1B1 at Y and A1B1 extended at X

alpha1 = alpha since BDB1E is concyclic

If < BAC = p, then < BXD = 90-alpha
So < B1YX = 180-2.alpha-2p = < MBC = 90- p since BM//BM1

Hence 2.alpha + p = 90

Now consider quadrilateral AGA1D

180-p + 180- theta + 2.alpha + p = 360

Therefore Theta = 2.alpha

Sumith Peitis
Moratuwa
Sri Lanka

3. For ease of understanding, read @ as Alpha
Extend AB to meet C1F at H
Observe that HB1D and HBE are similar => m(BEH)=m(BDB1)=@ -------------(1)
Let m(BAC)=A and extend BE to meet M1B1 at P
m(M1B1E)=180-m(MBE)=90+A=>m(M1C1B1)=m(M1B1C1)=m(EB1P)=90-A-@
Extend C1G to meet AB at Q => m(AQG)=m(M1C1B1)+m(AHC1)=90-A-@+90-@=180-A-2@
Therefore in Tr.AGQ, m(AGQ)=2@=m(CGC1) -----------------(2)