## Tuesday, August 30, 2016

### Geometry Problem 1253: Triangle, Orthocenter, Circle, Circumcircle, Altitude, Perpendicular, 90 Degree, Concurrency

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 1253.

1. Not knowing anything about this problem I tried to get the solution of P 1476 by the second point of the 3 circles C1HC2, A1HA2, B1HB2. But according to the great solution of P1476 by Pradyumna very easy join midpoints of segments FG, EJ, DK and according to the perpendicular bisector of the segment we get the point P

2. https://photos.app.goo.gl/RteRe7Yjg8Nqhbcm6

3. Here is another solution.
Let point Hb is reflection of point H across AC. It is easy to see that point Hb lies on circumcircle of triangle EHJ.
Similarly point Ha,Hc reflection of point H across BC and AB will lie on circumcircle of Triangle DHK and Triangle FGH respectively. .
Let these circumcircles of Triangle EHJ and DHK instersect at point P.
Angle HPHb= 2xAngle HEJ angle HPHa= 180 -2xAngle HDK
We get Angle HbPHa = Angle HPHb - Angle HPHa = 2x(Angle HEJ + HDK) -180 = 2x(180-C)- 180 = 180 - 2C
Since Angle HbBHa = 180 - 2C, hence Ha,Hb,B and P are concyclic. Ha,Hb and B lie on circumcircle of Triangle ABC hence P also lies on same circle.

Since Hc also lies on circum circle of Triangle ABC, We get Angle HcPHb = 2xAngle A
Angle HPHb=2x Angle HEJ, we get Angle HcPH = 2xA - 2xAngle HEJ = 2xAngle GFH
Since Angle HFHc is also 2xAngle GFH, points H,Hc,F and P are concyclic.
Hence Circumcircle of Triangle FGH also passes through P.