## Saturday, December 17, 2011

### Problem 704: Triangle, Orthocenter, Altitude, Midpoint, Perpendicular

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

Click the figure below to see the complete problem 704.

1. Quadrilateral BEHD is concylic . BH is the diameter of circle (EBHD).
Quadrilateral AEDC is concylic . AC is the diameter of circle (AEDC). These 2 circles intersect at E and D so line connected 2 centers will perpendicular to ED.
Peter Tran

2. AEDC is a cyclic quadrilateral, AC is the diameter and F is the circumcenter:then EF = DF.
EBDH is a cyclic quadrilateral,BH is the diameter and M is the circumcenter:then EM = DM.
Therefore EMDF is a Kite, then FM is perpendicular to DE.

3. Tr.s FEM & FDM are congruent SSS, hence FM is the perpendicular bisector of DE

Sumith Peiris
Moratuwa
Sri Lanka

4. Problem 704
Is ME=MD=BH/2, FE=FD=AC/2.Then the points F, M belong to the perpendicular bisector of ED.Therefore FM is perpendicular to DE.
MANOLOUDIS APOSTOLIS FROM GREECE