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### Existence of the Brocard Point

```Date: 06/06/2002 at 06:08:51
From: Razvan
Subject: Geometry - The Brocard Point

I have recently stumbled across a problem that asks to demonstrate
that the Brocard point exists in any triangle. I've searched the
Internet for the demonstration, but I couldn't find it.

Thanks,
Razvan
```

```
Date: 06/06/2002 at 06:58:49
From: Doctor Floor
Subject: Re: Geometry - The Brocard Point

Hi, Razvan,

Thanks for your question.

There are two Brocard points. One Brocard point is the point Om (in
most cases the Greek letter Omega is used) such that

angle(OmAB) = angle(OmBC) = angle(OmCA).

C

Om

A                 B

The second Brocard point is the point Om' such that

angle(Om'BA) = angle(Om'CB) = angle(Om'AC).

We shall prove that the first Brocard point exists.

Suppose that T is an interior point of the triangle such that

angle (TAB) = angle(TBC) = x.

Then

angle(TBA) = angle(B) - x,

and thus

angle(BTA) = 180 - angle(TBA) - angle(TAB)

= 180 - (angle(B)-x) - x

= 180 - angle(B)

So the locus of T lies on a circular arc, since angle(BTA) is fixed.
For reasons of ease we consider the complete circle. Let this circle
be Cc. In the same way the points U such that

angle(UBC) = angle(UCA)

yield a circle Ca.

The point of intersection Om of Cc and Ca, apart from B, is the first
Brocard point, because this point must satisfy

angle (OmAB) = angle (OmBC)

as well as

angle(OmBC) = angle (OmCA).

The circle Cc can be constructed in the following way: Intersect the
perpendicular bisector of AB with the line through B perpendicular to
BC, and let the point of intersection be X. Then

angle(BAX) = angle(XBA)

=90 - B,

and thus

angle(AXB) = 2B,

so the circle with center X through A and B is indeed Cc.

Mathworld - Brocard points
http://mathworld.wolfram.com/BrocardPoints.html

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 06/11/2002 at 05:36:23
From: Razvan
Subject: Geometry - The Brocard Point

Hi, Doctor Floor,

Thank you very much for the demonstration, but I would like to ask
you more about it.

Why is it that if the angle(BTA) is fixed, the locus of T is a
circular arc?

The second question is about the X point. Why is it that if the
angle(AXB) = 2B, the X point is the center of Cc?

Thanks,
Razvan
```

```
Date: 06/11/2002 at 06:59:58
From: Doctor Floor
Subject: Re: Geometry - The Brocard Point

Hi,

Thanks for your question.

Both have to do with inscribed and central angles of a circle.  Let me
therefore point you to the following message from the Dr. Math
archive:

Inscribed Angle Theorem
http://mathforum.org/library/drmath/view/55073.html

It tells that two inscribed angles on the same arc of a circle are
half the measure of the central angle on that same arc. That should

For your first question, we need that the locus of angles at the same
side of a segment making congruent angles on that segment is a
circular arc.

To see that, we suppose that we have an angle from C on segment AB.
We draw the circumcircle of ABC:

Of course the circular arc AB containing C is part of the locus, we
know this from the inscribed angle theorem.

Now suppose that E is a point in the exterior of the circle, and also
part of the locus. Then we let line AE intersect the circumcircle
again in point D. We know that

angle ADB = angle ACB,

but we also know that

angle AEB = 180 - angle EDB - angle DBE

= angle ADB - anlge DBE.

So angle AEB is not congruent to angle ADB = angle ACB. We have a

In the same way we show that an interior point F of the circle cannot
be part of the locus.

If you need more help, just write back.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 06/14/2002 at 04:01:46
From: Razvan
Subject: Geometry - The Brocard Point

Hi,

I still have a question about the X point. The X point is the center
of the Cc, T is on Cc, and angle(AXB) = 2B.

Then, shouldn't

angle(BTA) = angle(B)

because angle(AXB) is a center angle and angle(BTA) is inscribed on
the same arc as angle(AXB)? Instead

angle(BTA) = 180 - angle(B).

Thanks,
Razvan
```

```
Date: 06/14/2002 at 08:14:32
From: Doctor Floor
Subject: Re: Geometry - The Brocard Point

Hi, Razvan,

Thanks for your question.

Angle(BTA) is an inscribed angle on the opposite arc on which
angle(AXB) is a central angle. Together these arcs are 360 degrees.
You can see that from the fact the order of A and B in BTA and AXB is
opposite.

If you have more questions, just write back.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Triangles and Other Polygons
High School Triangles and Other Polygons

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