## Friday, March 10, 2023

### Geometry Problem 1524: Unlock the Mystery of Parallelograms: Discover the Length of Segment between the Intersecting Angle Bisectors. Difficulty Level: High School.

Geometry Problem 1524. Post your solution in the comment box below.
Level: Mathematics Education, K-12 School, Honors Geometry, College.

Details: Click on the figure below. More Details

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1. Let AE,DF extended meet at X and let BE,CF extended meet at Y.
Let BE = DF = p and AE = FC = q. Let FY = u and EY = v

< ABC & < BAD are supplementary and so < ABE & < BAE are complementary
Hence EX FY is a Rectangle

From similar triangles ABE & BYC,
(p+v)/p = (q+u)/q = 10/6
So u = 2q/3 and v = 2p/3

Applying Pythagoras to Triangles ABE & EYF,
EF^2 = u^2 + v^2 = (2/3)^2. (p^2 + q^2) = (2/3)^2. (6^2) = (4/9)*36 = 16
Therefore EF = 4

Sumith Peiris
Moratuwa
Sri Lanka

2. Extend DF to P => BP=10 - 6 = EF

1. How is BP = EF?

2. On what basis is BP = EF?

3. ( BE = PF, & BE // PF ) => BPFE parallelogram

3. Dear Peter - could you explain how you concluded that points M,E,F,N are collinear which is essential for your proof to be valid?

Regards

Sumith

4. Since M and N are midpoints of AB and CD => MN//BC
ME//BC and NF //BC => M,E, F ,N are collinear

1. 