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

Labels:
angle,
congruence,
isosceles,
median,
perpendicular,
triangle

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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)