Monday, April 1, 2013

Problem 867: Isosceles Triangle, Median, Perpendicular, Angle, Congruence

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

Click the figure below to open the complete problem 867.

Online Geometry Problem 867: Isosceles Triangle, Median, Perpendicular, Angle, Congruence
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4 comments:

  1. AjitApril 6, 2013 at 2:29 AM

    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

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  2. Sumith PeirisNovember 9, 2015 at 9:03 PM

    Let bisector of < ABC meet AD at F and AC at G. Let < CAD = < EBD = @

    ABEG 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

    ReplyDelete
  3. Stan FulgerJuly 6, 2020 at 10:36 AM

    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.

    Best regards,
    Stan Fulger

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  4. c.t.e.o.September 30, 2020 at 12:52 PM

    Draw BLM the median of ABC. => BLD isosceles => LE=1 and LD=2
    But LD=1/3(x+1)

    ReplyDelete

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