Geometry Problem

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

Click the figure below to see the complete problem 704.

## Saturday, December 17, 2011

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

Labels:
altitude,
midpoint,
orthocenter,
perpendicular,
triangle

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Quadrilateral BEHD is concylic . BH is the diameter of circle (EBHD).

ReplyDeleteQuadrilateral 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

AEDC is a cyclic quadrilateral, AC is the diameter and F is the circumcenter:then EF = DF.

ReplyDeleteEBDH 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.

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

ReplyDeleteSumith Peiris

Moratuwa

Sri Lanka

Problem 704

ReplyDeleteIs 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