## Thursday, May 10, 2012

### Problem 751: Parallelogram, Congruence, Intersecting Lines, Angle Bisector

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

This entry contributed by Ajit Athle.
Click the figure below to see the problem 751 details.

1. http://img204.imageshack.us/img204/4091/problem751.png

Extend BG to cut CD at H .
Extend CG to cut BA at P ( see sketch)
Let AB=b , BC=a, ED=BF=x
∆ (BCH) similar to ∆ ( EDH) so HC/a=HD/x=(HC-HD)/(a-x)= b/(a-x)
So HD=bx/(a-x) and CH=ab/(a-x)
∆ (FBG) similar to ∆ (DGH) and ∆ (PBG) similar to ∆ (CHG)
So BF/DH=BG/GH=BP/CH=(a-x)/b
And BP=CH (a-x)/b = ab/(a-x). (a-x)/b= a
So BP=BC=a and triangle CBP is isosceles
∠ (BCP)= ∠ (BPC)= ∠ (PCD)….(Alternate angles)
CG bisect ∠ (BCD)

2. Extend BE and CD to meet at H.
BC/CH = ED/DH = BF/DH =BG/GH
Hence CG bisects angle BCH

3. Proof:
Through G construct PQ // AD so that P on AB and Q on DC
Through G construct RS // AB so that R on AD and S on BC
Triange GPB and GRE are similar
PB : PG = RG : RE
Triangle GPF and DRG are similar
PF : PG = RG : RD
PF : PB = RE : RD
(PB – BF) : PB = (RD – DE) : RD
BF : PB = DE : RD
BF = DE
PB = RD
GSCQ is a rhombus
CG is the bisector of angle BCD

4. Dear Pravin,

Sorry that I don't understand your proof.
Why a pair of triangles with two sides ratios equal them their are similar?

Kent

5. Triangles BCH and EDH are similar-> first two equalities;BF=ED -> 2 and 3rd equality;BFG and GDH are similar-> last two equalities. And BC/CH=BG/GH is property of bisector,hence GC is bisector of angle C. I solved problem completely indetical:)

6. http://www.mathematica.gr/forum/viewtopic.php?f=22&t=27407&p=134063