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.

Online Geometry Problem 751: Parallelogram, Congruence, Intersecting Lines, Angle Bisector.



    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?


  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:)