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An Euler Circle Proof

Date: 03/26/99 at 13:02:48
From: Bernadette Gural
Subject: The Euler Circle

I am having trouble proving the following result:

The distance, d, between the circumcenter O and the incenter, I, of a 
triangle is given by the relation: d^2 = R(R-2r), where R is the 
circumradius and r is the inradius.

I know that the nine points are consisted of the following:
J, K, L - the Euler points
D, E, F - the feet of ha, hb, hc of triangle ABC
A', B', C' - the midpoints of triangle ABC

I guess I am having trouble with the incenter (not sure what exactly 
an incenter is) and also with the inradius.

Is there any help you can give me to prove the above result? I 
appreciate both your time, effort and patience in dealing with yet 
another stumped geometry student.

Date: 03/26/99 at 14:41:52
From: Doctor Floor
Subject: Re: The Euler Circle

First: the question you pose is a theorem by Euler (1765, an earlier 
statement is by Chapple 1746), but it is not about the nine point 

Second: the incenter is the point of intersection of the three 
(internal) angle bisectors in a triangle. It is also the center of the 
circle inscribed in the triangle (= incircle). That means that this 
circle is inside the triangle and is tangent to the three sides.

Third: The proof.

Let ABC be the triangle, O its circumcenter and I its incenter. Let R 
be the radius of the circumcircle and r of the incircle. Let E be the 
point where the incircle meets BC, and let D and D' be the midpoints 
of arcs AB over the circumcircle. Finally, let d = OI.

Here is a picture of the situation:


We observe that angle ACD = 0.5* angle C, since D bisects arc AB. As a 
consequence, I is on segment CD. It is also easy to see that DD' is a 
diameter of the circumcircle.

Note that angle DBI = angle DBA + angle ABI
                                = angle DCA + 0.5*angle B (cuts off 
                                                           same arc)
                                = angle ICA + 0.5*angle B
                                = 0.5*angle C + 0.5*angle B

Note also that angle BDI = angle BDC
                                = angle A (cuts off same arc)

So angle BID = 180 - angle A - 0.5*angle C - 0.5*angle B
             = 0.5*angle C + 0.5*angle B

So, triangle BID is isosceles, and DB = DI.

It is not difficult to see that triangles D'BD and CEI are similar, 
since angle DD'B = angle DCB = angle ICE and angle D'BD = angle CEI = 
90 degrees. We can use this to get the following ratio:

    DD':DB = IC:IE

From which we can deduce:

    DD'*IE = IC*DB [DD' = 2R, IE = r and DB = DI]
    2R*r = IC*ID

But IC*ID is equal to the power of incenter I w.r.t. the circumcircle. 
As a consequence IC*ID = R^2 - OI^2 or:

    2Rr = R^2 - d^2 
    d^2 = R^2 - 2Rr
    d^2 = R(R - 2r)

And the statement is proven.

- Doctor Floor, The Math Forum
Associated Topics:
High School Conic Sections/Circles
High School Geometry

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