## Wednesday, July 3, 2013

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

Geometry Problem. GeoGebra, HTML5 Animation for iPad and Nexus.

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

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

#### 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.
http://img856.imageshack.us/img856/203/5d0.png
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