Touching Circles and Concurrency

The applet below illustrates the following problem that has been sent by Ionut Onisor, Bucharest, Romania:

Let AHa, BHb and CHc be the altitudes of ΔABC. Draw a circle through Hb and Hc tangent to circumcircle. Let A' be the point of tangency and B' and C' similarly defined. Prove that lines AA', BB' and CC' have a common point.

As you can check with the applet, the statement requires qualifications: first, it may only be true for acute triangles. Second, there may be two circles through Hb and Hc tangent to circumcircle. The statement only holds when we pick the circle such that triangles AHbHc and A'HbHc have different orientation; in other words, when line HbHc separates A and A'.

In addition, the lines A'Ha, B'Hb, C'Hc are also concurrent and the point of concurrency appears to lie on the Euler line of ΔABC.

 

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

Solution

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Copyright © 1996-2017 Alexander Bogomolny

 

 

This applet requires Sun's Java VM 2 which your browser may perceive as a popup. Which it is not. If you want to see the applet work, visit Sun's website at http://www.java.com/en/download/index.jsp, download and install Java VM and enjoy the applet.


What if applet does not run?

... to be continued ...

|Activities| |Contact| |Front page| |Contents| |Geometry| |Store|

Copyright © 1996-2017 Alexander Bogomolny

 61229103

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