Geometry Stumper

[jwhsu]

(08-14-2018 08:35 PM)Thomas Puettmann Wrote:  

(08-14-2018 06:58 PM)Albert Chan Wrote:  What does midpoint of a circle mean ?

center, sorry.

A standard theorem from elementary geometry says that all three (interior) angular bisectors meet at one point, namely, the center of the inscribed circle.

How can one proof this? Imagine an arbitrary point on an angular bisector. Keeping this point as the center start blowing this point into a larger and larger circle. For one and only one radius the circle will touch both rays of the angle simultaneously.

Now, imagine you have two (interior) angular bisectors in a triangle. They meet at one and only one point. The circle just described now touches all three sides of the triangle. This circle is called inscribed circle of the triangle. It follows immediately that the third angular bisector must pass through the center of this circle.


To conclude the proof of your original problem you do the same at D outside of the triangle ABC. Just watch out what you get for free.

Is this basically using an excircle to solve the problem? (I had to look it up, it has been a long while since I took geometry). From what I recall, one of its properties is the center of an excircle of a side of a triangle is the intersection of the two external angle biscectors and the extension of the opposite interior angle bisector of a triangle.

For this problem, if you extend sides AC and BC, the angle opposite ACB is identical. This means the angles adjacent to ACB are also identical. Using the properties of excircles, point D is the center of an excircle for line AC and segment CD bisects the two exernal angles of the triangle on that side. Similar logic goes for line BC and its corresponding excircle. This means the angles around point C are 2*ACB + 2*ACD + 2* BCE = 360°. Dividing in half gives us ACB + ACD + BCE = 180°, which means that segment DCE is a straight line.