SECTION 1.3 Circle Geometry 35
- Prove Van Schooten’s theorem.
Namely, let 4 ABC be an equilat-
eral triangle, and letC be the cir-
cumscribed circle. Let M ∈ C
be a point on the shorter arcBC ̆.
Show that AM = BM + CM.
(Hint: Construct the point Dsub-
ject to AM =DM and show that
4 ABM∼= 4 ACD.) - The figure to the right shows the
triangle 4 ABC inscribed in a cir-
cle. The tangent to the circle at
the vertex A meets the line (BC)
atD, the tangent to the circle atB
meets the line (AC) at E, and the
tangent to the circle atCmeets the
line (AB) at F. Show that D, E,
and F are colinear. (Hint: note
that 4 ACD∼4BAD(why?) and
from this you can conclude that
DB
DC =
(AB
AC) 2. How does this help?)
1.3.3 Cyclic quadrilaterals and Ptolemy’s theorem
As we have already seen, any triangle can be incribed in a circle; this
circle will have center at the circumcenter of the given triangle. It is
then natural to ask whether the same can be said for arbitrary polygons.
However, a moment’s though reveals that this is, in general false even
for quadrilaterals. A quadrilateral that can be incribed in a circle is
called acyclic quadrilateral.