Interactive step-by-step animation using GeoGebra. Post your solution in the comment box below.

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

Details: Click on the figure below.

## Tuesday, April 28, 2020

### Dynamic Geometry 1476: Droz-Farny Line Theorem, Triangle, Orthocenter, Perpendicular, Collinear Midpoints

Labels:
altitude,
classical theorem,
collinear,
Droz-Farny,
GeoGebra,
ipad,
ipadpro,
line,
orthocenter,
perpendicular,
triangle

Subscribe to:
Post Comments (Atom)

Let the mid points of A1A2,B1B2 and C1C2 are Ma,Mb and Mc.

ReplyDeleteDraw perpendicular MaD to BH and MbE to AH.

We get MaD/MbE=BMa/AMb ......(1)

Let Angle HA2A1=x and Angle HB1B2=y,

Then Angle MbHE=Angle MbHA1+ Angle A1HE=x+y

Similarly Angle DHMa = Angle DHA2 + Angle MaHA2=x+y

Hence Triangle HMbE is similar to triangle HMaD,

We get MaD/MbE=HMa/HMb,using eq 1, we have BMa/AMb=HMa/HMb

Similarly we get CMb/BMc=HMb/HMc, and AMc/CMa=HMc/HMa,

We get (AMc/BMc)x(BMa/CMa)x(CMb/AMb)=1, hence by Menelaus Theorem Ma,Mb and Mc must be collinear.

Using results of Problem 1253, circumcircles of Triangle HA1A2, HB1B2,HC1C2 and ABC, pass through point P. Since HP is common chord of 3 circles, its centers will lie on perpendicular bisectors of HP. Since Ma,Mb and Mc are also center of these triangles.

ReplyDeleteMa,Mb and Mc will be collinear.