Geometry Problem
Level: Mathematics Education, High School, Honors Geometry, College.
Click the figure below to open the complete problem 867.
Monday, April 1, 2013
Problem 867: Isosceles Triangle, Median, Perpendicular, Angle, Congruence
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Draw CF perpendicular to Ad extended. /_FCD=/_DBE=/_DAC; hence FC is tangent to the circum-circle of tr. ADC and thus CF²=BE²=1*(x+2) since FD=DE=1. Now AB²=BE²+x², AB=2*BD and BD²=1+BE² give us: x²+2+x=4(x+2)+4 which in turn gives x=5
ReplyDeleteLet bisector of < ABC meet AD at F and AC at G. Let < CAD = < EBD = @
ReplyDeleteABEG is cyclic with diameter AB so < FBE = @. Hence FE = 1 and AF = x-1
Since BF is an angle bisector of Tr. ABD, AB/BD = (x-1)/2 from which x = 5 since AB/BD = 2
Sumith Peiris
Moratuwa
Sri Lanka
BG, reflection of BD about BE is median of tr. ABC, thus G is the centroid and AF=2DF=4, hence AE=AG+GE=4+1=5.
ReplyDeleteBest regards,
Stan Fulger
Draw BLM the median of ABC. => BLD isosceles => LE=1 and LD=2
ReplyDeleteBut LD=1/3(x+1)