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Geometry ProblemLevel: Mathematics Education, High School, Honors Geometry, College.Click the figure below to see the problem 710 details.
1. AC bisect BD , CO is a median of triangle BCD So CD will pass through center of gravity N of triangle BCD and A,O,P,N, C are collinear and EF//BD2. Solution of problem 709 showed that AG/GE=AH/AF = 4 so GH//EF//BDIn triangle BCD, since O is midpoint of BD so P is the midpoint of MQIn triangle BND, since O is midpoint of BD so P is the midpoint of GHSo PG=PH=HQ=GM3. see item 2 above4. in quadrilateral MBDR, BD//MR and BM//DR so MQDR is a parallelogram and BD=MR
Additional clarifications for question 2 ( MG=GP=PH=HQ=QR )Let BD=x and BD cut AE and AF at S and T1. In tri. ABC , S is the centroid so BS=2/3 BO=x/3Similarly DT=x/32. In tri. BND , since NO is a median so P is the midpoint of GH Similarly in tri. BCD , P is the midpoint of MQ3. Tri BGS~Tri. FGE so BG/GF=BS/EF=2/3 BG/BF=2/5 GM=x/5FG/GB=3/2 FG/FB=3/5 = GH/BT GH=3/5* 2x/3=2x/5 PQ=1/2GH=x/5=PHPQ=PM=GM+1/2GP=2x/5 HQ=PQ-1/GH=x/54. BDRM is a parallelogram MR=BD=x QR=MR-MQ=x/5So MG=GP=PH=HQ=QR=x/5