## Wednesday, February 15, 2012

### Problem 729: Triangle, Altitude, Orthocenter, Circumcircle, Perpendicular, Midpoint

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

Click the figure below to see the problem 729 details.

1. http://img713.imageshack.us/img713/4057/problem729.png

Let BH cut the circle at L . ( see picture)
Let P is the projection of D over AC and Q is the point of symmetry of D over AC
We have NH= NL ( property of orthocenter)
And E, F and P are collinear ( Simson line)
HLQD is isosceles trapezoid ==> N, S , P are collinear
HSL is a isosceles triangle => ∠ (NHS)= ∠ (NLS)= ∠ (SQP) ( line HL // DQ)
Quadrilateral DFPC is cyclic => ∠ (FPD)= ∠ (FCD) ( both angles face arc FD)
But ∠ (FCD)= ∠ (BLD)= ∠ (NLS) ( angles face same arc BD)
So∠ (FPD)= ∠ (NLS)= ∠ (SQP) => line EFP // line HSQ
Since P is the midpoint of DQ so M will be the midpoint of HD

Let H’ be the orthocenter of ∆ADC
Let K, J be the feet of the ⊥s BH, DH’ on AC
Let KB, JE (both extended) meet at X
Let DJ(extended)meet circumcircle at Y
Join BY, DC. Also join XC, HY
We prove XHJD is a parallelogram
which would imply that the diagonals
XJ, HD bisect each other.
BH = DH’ (∵ each = 2 OZ, Z midpoint of AC)
C, D, F, J are cyclic.
∴ ∠FJD = ∠FCD = ∠BYD and so XJ ∥ BY
=> XBYJ is a parallelogram (∵XB, JY each ⊥ AC)
=> XB = JY = H’J (Property of orthocenter)
∴ XH = XB + BH = H’J + DH’ = DJ
Hence XHJD is a parallelogram, as required.

3. Refer Problem 729
http://imageshack.us/photo/my-images/338/729gg.pdf/