## Wednesday, September 4, 2013

### Problem 917: Triangle, Congruence, Midpoints, Angle Bisector, Perpendicular

Geometry Problem
Post your solution in the comment box below

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

Click the figure below to see the complete problem 917.

1. Property # 1 of a parallelogram: angle bisectors of 2 opposite angles are parallel.
Let 2 angle bisectors of a parallelogram ABCD cut opposite sides at E and F .
Triangles ABF and CDE are congruence per case ASA => DE=BF and BE=DF => DE //BF

Draw additional lines per sketch and let L is the midpoint of AC
We have NL= ½ AF and NL//AF , ML=1/2 CE and ML//CE
BPLQ is a parallelogram
Since AF=CE => ML=NL => triangle MLN is isosceles
Draw angle bisector LR of angle MLN => LR ⊥ MN
LR // BD per property #1 of parallelogram BPLQ => BD ⊥ MN

2. Connect E with F. Let P be the mid-point of EF, and Q the mid-point of AC.
PM is parallel to AF. PM is equal to 1/2 of AF. NQ is parallel to AF. NQ is equal to 1/2 of AF. PN is parallel to EC. PN is equal to 1/2 of EC. MQ is parallel to EC. MQ is equal to 1/2 of EC. PMQN is a rombus, and its diagonals are perbendicular to each other. PQ is bisector of <MPN. BD is parallel to PQ, dhe PQ is perbendicular to MN, and from here we have BD is perbendicular to MN.

Erina-NJ

1. Note to Erina
Refer to your solution in the paragraph below ,
" PMQN is a rombus, and its diagonals are perbendicular to each other. PQ is bisector of <MPN. BD is parallel to PQ"
It is not clear to me how do you have "BD is parallel to PQ " . please explain.

3. To Peter Tran:

PM is parallel to AB, PN is parallel to BC. <ABC dhe <MPN are angles with parallel sides. This way, their bisectors(BD dhe PQ) are parallel or they fall in the same line.

Erina-NJ

4. Q-середина AC, C' пересечение MN и AB, A' пересечение MN и BC. QM параллельна EC, QN параллельна AF, и QM=EC/2=AF/2=QN. <BC'M=<MNQ=<NMQ=<NA'B, то есть треугольник BA'C' равнобедренный.

5. Let Q be midpoint of AC, C' intersection of MN with AB, and A' intersection of MN with BC. QM parallel to EC, QN parallel to AF, and QM=EC/2=AF/2=QN. <BC'M=<MNQ=<NMQ=<NA'B, making triangle A'BC' isosceles.