Wednesday, July 3, 2013

Problem 895: Triangle, Incircle, Tangency Points, Parallel lines, Collinear Points

Geometry Problem. GeoGebra, HTML5 Animation for iPad and Nexus.
Post your solution in the comments box below.

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

Click the figure below to see the dynamic geometry demonstration of problem 895.

Online Geometry Problem 894: Triangle, Angle, 60 degrees, Incenter, Midpoint, Parallel Lines. GeoGebra, HTML5 Animation for iPad and more tablets

1 comment:

  1. Lemma 1: In trapezoid ABCD, let E is the intersection of diagonals AC and DB and F is the point of intersection of AD and BC then EF cut upper and lower bases at midpoints of the bases.
    See problem 714 for full demonstration.
    Let L and M are midpoints of GD and EH
    1. Apply Lemma 1 in trapezoid GDHE , FH will cut GD and EH at midpoints of DG and EH
    .=> F, L,K and M are collinear
    2. From C draw a line parallel to AB . This line will cut EF and FD at E’ and D’ ( see sketch)
    We have ∆(FAE) similar to ∆ (E’CE) ( case AA)
    So ∆ (ECE’ )is isosceles and CE’=CE
    Similarly ∆(FBD) similar to ∆(D’CD) => ∆( D’CD) is isosceles => CD=CD’
    But CE=CD so CE’=CD’ => C is the midpoint of E’D’
    3. Let line FLKM cut E’D’ at C’
    With similar triangles we get C’E’/ME= FC’/FM=C’D’/MH
    But ME=MH so C’E’=C’D’ => C’ is midpoint of E’D’ and C’ coincide to C

    So F, K and C are collinear