1. Suppose that two lines through a point P are tangent to a circle at points A and B. If the circle has center O and radius R, and Q is the intersection point of the lines OP and AB, then (OP)(OQ) = R^2.
In the figure OAPB, R is the radius of the circle. From definition, if a line is tangent to a circle at a given point then the radius is perpendicular to that line at that point. With this in mind, angle OAP and OBP are right angles. Given OA = OB = radius of the circle, and both triangle OAP and OBP share side OP the by side angle side congruence then triangle OAP is congruent to OBP. By using congruence theorem, then side AP is equal to side BP. Now if that is so then by using isosceles triangle theorem, angel OAQ is equal to angle OBQ. But note that OPA is congruent to OPB which means that angle AOP is congruent to angle BOP. Now consider triangles AOQ and BOQ, given angle OAQ = OBQ, BOQ = AOQ, then by definition of a triangle, angle AQO = BQO which are supplementary by straight angle theorem. Now if two angles are supplementary and equal, then they are right angles. Therefore we can conclude that angle AQO and OQB are both right angles. By side angle side similarity, triangle AOP is similar to AOQ. On the other hand, triangle OQB is similar to OBP.
By using similarity theorem, then OP/R = R/OQ
by cross multiplication, then OP*OQ = R^2
qed.for the gsp sketch click
here
2. If two lines
through a point P meet a circle at points A, A' and B, B', respectively,
then (PA)(PA') = (PB)(PB'). (Explanation: One line meets
the circle at two points A and A', and the other line meets the
circle at two points B and B'.)
The image created by gsp below shows that the relationship is true. To play with the gsp file click here.
In the figure above, it
can be noted that AA'B'B is cyclic quadrilateral. By definition
opposite angle of cyclic quadrilateral are supplementary. Which
means angle BAA' + angle BB'A' and angle B'A'A + angle ABB' are
supplementary. From straight angle theorem, angle y, x and a area
supplementary. Which means that angle x + y = (180 - a). Given
angle A'B'B and A'AB are supplementary and angle A'AB = 180 -
a, then angle BB'A' = a. With the same point, angle ABP = angle
B'A'A. Given both triangles share angle APB, then by angle, angle
angle similarity, triangle ABP is similar to triangle A'PB'. Hence
by using similarity theorem, PA/PB = PB'/PA' by cross multiplication,
it becomes PA*PA' = PB*PB'
3. The feet of the perpendiculars from a point to the
sides of a triangle are collinear if and only if the point lies
on the circumcircle. (Explanation: The foot of the perpendicular
from a point to a line is the intersection of the line with the
perpendicular line through the point.)
It looks like the points P, J and I lie on the same line. The strategy here is to show that angle PJI = 180, by straight angle theorem, points P, J and I are collinear. Given the constructed lines ( yellow) are perpendicular to the sides of triangle ABC, then angle CJD = 90. Construct a line parallel to JD through the center and point S. Let angle DJP = x and angle IJM = y.
Hence by corresponding angle theorem, angle PJD = angle PES = x
Consider triangle EJM by triangle sum theorem, x + y + 90 = 180 ..(1)
Now in a line which contains P, J and I, then from (1) x + 90 + y = 180 then by straight angle theorem, P,J and I are collinear. Click here for the gsp file