Geometry Problem. Post your solution in the comment box below.

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

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## Thursday, January 14, 2016

### Geometry Problem 1179: Triangle, Centroid, Transversal, Sum of Ratios, Sides

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http://s9.postimg.org/71ehki32n/pro_1179.png

ReplyDeleteDraw median BM and lines AL//DE and CN // DE

Note that triangles ALM and CNM are congruence… ( case ASA)

We have a1/a2= GL/GB=(GM-ML)/GB= ½ - ML/ GB …. (1)

C1/c2= GN/GB= ( GM+MN)/GB = ½ + MN/GB …. (2)

Add ( 1) to (2) and note that ML=MN we get the result.

MANOLOUDIS APOSTOLIS 4 HIGH SCHOOL OF KORYDALLOS PIRAEUS GREECE

ReplyDeleteBG intersecting AC M is AM=MC and BG=2.MG ,si DE intersecting AC K .

Apply the theorem Menelaus tri ABM and tri BMC intersecting by KDGE.

Therefore AK/KM.MG/GB.BD/DA=1 or DA/DB=AK/2.KM and MK/KC.EC/EB.BG/GM=1 or

EC/EB=KC/2.MK. Therefore DA/DB+EC/EB=AK/2.MK+KC/2.MK=(AK+KC)/2.MK=(MK-AM+KM+MC)/2.MK=2.MK/2.MK=1.