## Thursday, April 14, 2016

### Geometry Problem 1207: Triangle, Circle, Incenter, Circumcenter, Excenter, Circumradius, Perpendicular

Geometry Problem. Post your solution in the comment box below.
Level: Mathematics Education, High School, Honors Geometry, College.

Click the figure below to view more details of problem 1207. 1. http://s22.postimg.org/nia3yt2fl/pro_1207.png

Let L, M, N , P are points as shown on the sketch
Observe that AL= NC= half perimeter of triangle ABC- BC
P is the midpoint of arc AC => M is the midpoint of AC and LN
In trapezoid ILND , MO is the mid-base => O is the midpoint of ID
Triangle IOP similar to triangle IDE ..( case AA)
Since O is the midpoint of ID so DE= 2 x OP= 2.R

2. Lets assume touch point of incircle and excircle are F and G, and Midpoint of AC is M. It is easy to see that FM=GM=(a-c)/2 Also IF,OM and DG are parallel to each other ( all are perpendicular to AC), hence O is midpoint of ID.
B, I and E are collinear, join BE and let it intersects circumcircle at point H, and IH=HE, thus H is midpoint of IE.
Consider triangle IED, O is midpoint of ID and H is midpoint of IE,
Hence DE=2*OH, since OH=R, DE=2R.

3. Join BIE. Let it cut circle(O)at M. Angle ECI is a right angle.
M is the midpoint of arc AMC. So OM bisects AC at right angles.
We are done if we can show O is the midpoint of ID.
Let X, Y, Z be the projections of I. O. D on AC respectively.
It is easy to see that XY = b/2 - (s -a) = (c -a)/2 = YZ.
IX, OY, DZ being //, OYM is a midline // to DE in Triangle IDE.
Hence DE = 2 OM = 2R.
4. 