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.

## Thursday, May 10, 2012

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

Labels:
angle bisector,
congr,
parallelogram,
similarity

Subscribe to:
Post Comments (Atom)

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

ReplyDeleteExtend 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)

Extend BE and CD to meet at H.

ReplyDeleteBC/CH = ED/DH = BF/DH =BG/GH

Hence CG bisects angle BCH

Excellent!

DeleteProof:

ReplyDeleteThrough 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

Dear Pravin,

ReplyDeleteSorry that I don't understand your proof.

Why a pair of triangles with two sides ratios equal them their are similar?

Kent

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:)

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

ReplyDelete