Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.
Details: Click on the figure below.
Saturday, February 10, 2018
Geometry Problem 1355: Triangle, Midpoint, Median, Circle, Chord, Equal Product of measure of segments
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https://photos.app.goo.gl/h4PcQ74of1KJy3eh2
ReplyDeleteDraw AP//DE and CN//DE
See sketch for positions of points P,Q, N
Triangle BAP is isosceles => BP=BA
Triangles AQM and CNM are congruent ( case ASA) => AQ=NC
Since DE//AP => GE/DG=PQ/AQ=PQ/NC
Triangle BPQ similar to BCN => PQ/NC= BP/BC=AB/BC => GE/DG=AB/BC
Or GE . BC= DG . AB
Draw perpendiculars h1 and h2 from G to AB and BC
ReplyDeleteSince S (ABG) = S (CBG),
AB.h1 = BC.h2..... (1)
Since BD = BE,
GE.h1 = DG.h2..... (2)
Divide (1) by (2) and the result follows
Sumith Peiris
Moratuwa
Sri Lanka
Using areas it is straight forward.
ReplyDeleteBest regards
Let BM split angle ABC into angles x and y.
ReplyDeleteArea of triangle ABM = Area of triangleBCM.
So ½ AB.MB.sin x= ½ BC.MB.sin y,
Implies AB/BC = sin y/sin x .
Next EG/DG = (BGE)/(BGD)
= ½ BE.BG sin y / ½ BD.BG sin x
= sin y/sin x
Hence AB/BC = EG/DG,
AB.DG = BC.EG
Vijaya Prasad Nalluri (Pravin)
Rajahmundry - INDIA