Pascal's theorem is a very useful theorem in Olympiad geometry to prove the collinearity of three intersections among six points on a circle.
The theorem states as follows:
Given 6 points (which can be coincident) on the circumference of a circle labelled and in that order around the circle, the intersections of and , and , and and are collinear.
There are many different ways to prove this theorem, but an easy way is to use Menelaus' theorem.
Let be the intersection of and let be the intersection of and and let be the intersection of and We will prove these three points are collinear.
Let be the intersection of and let be the intersection of and and let be the intersection of and By Menelaus in and line , we have
By Menelaus in and line , we have
By Menelaus in and line , we have
Multiplying these out, we get
Upon rearranging, we get
Notice that by power of a point, we have
Thus, the product above simplifies as
So, by Menelaus, and are collinear.
Consider a circle internally tangent to the circumcircle of triangle and also tangent to sides and at and respectively. Show that passes through the incenter of
At a glance, we can see that we want to be the points of collinearity, but this may not seem obvious at first, especially since there are only 3 points on the circumcircle. However, if we break this question down into small, manageable chunks, we can break through with solving this.
Let be the point of tangency between the circumcircle and the smaller circle (this circle is called the mixtilinear circle). We extend to meet the circumcircle again at . We claim that , and we prove it as follows:
Let intersect the mixtilinear circle again at , and let be the centre of the mixtilinear circle. By alternate segment theorem, is equal to the angle made by and the tangent, which is equal to , so . We will call and . The following angle chase is not too hard and is left as an exercise for the reader. Angle chasing results in , and , from which we get and , so . Thus, is isosceles, so .
Thus, our claim is proven. From this statement, we can invoke the following property:
If is a point on the circumcircle of on the minor arc if and only if is the angle bisector of .
From this property, we get is the angle bisector of . Similarly, if is extended to , , so is the angle bisector of . Now we can finally apply Pascal's theorem. By Pascal's theorem in (following the line segments in that order and cycling back), we get , and as the collinear intersection points, so we are done.
Prove the fundamental theorems of poles and polars:
Given a cyclic quadrilateral , if the intersection of and is , the intersection of and is , and the intersection of and is , prove the polar of passes through and .
Firstly, here are some definitions. Let be the center of a circle with radius , and an arbitrary point, which we will call the pole. Let be the point on (possibly extended) such that . The line through perpendicular to is known as the polar of . A unique property (left to be proven by the reader) of poles and polars is that if is inside the circle, then is the intersection of the tangents from the endpoints of the chord with midpoint .
The way to solve this problem is with Pascal's. Note that Pascal's can be applied even if two or more points are coincident. Let us consider Pascal's in hexagon . Then, , (the line through coincident points would be a tangent to the circle) and are collinear. Similarly, if we use Pascal's in , we have , and . Thus, , , and are collinear. A similar argument shows , , and are collinear. From this, using the fact that is a tangent, etc., we get that is parallel to , from which we can gather is a pole of .