These are the proofs of certain theorems and claims that will be useful when reading the proof of the nine point circle.
Thale's Theorem
Theorem: A triangle ABC with circle center O and with side AB a diameter of the circle, for any point C on the circle, angle ACB is a right angle.
Proof: OA is congruent to OB because they are all radii of the circle, so triangle OBA and triangle OBC are both isosceles. Therefore measure angle OBC=measure angle OCB, and measure angle OAB=measure angle OBA. If we call angle OAB a and angle OBC b, then the three angles of triangle ABC are a, (a+b), and b. So a+(a+b)+b=180, 2a+2b=180, 2(a+b)=180, so a+b=90. (Wikipedia, "Thales' Theorem")
Theorem: A triangle ABC with circle center O and with side AB a diameter of the circle, for any point C on the circle, angle ACB is a right angle.
Proof: OA is congruent to OB because they are all radii of the circle, so triangle OBA and triangle OBC are both isosceles. Therefore measure angle OBC=measure angle OCB, and measure angle OAB=measure angle OBA. If we call angle OAB a and angle OBC b, then the three angles of triangle ABC are a, (a+b), and b. So a+(a+b)+b=180, 2a+2b=180, 2(a+b)=180, so a+b=90. (Wikipedia, "Thales' Theorem")
Euler's Line
Theorem: There exists a line that passes through the centroid, circumcenter, and orthocenter of a triangle.
Proof: First there are some constructions that need to be done. We begin by creating a triangle ABC, then finding the midpoints of ead segment, AB, BC, and CA which are D, F, and E respectively. Then we create a smaller triangle DEF. We will find the orthocenter of triangle DEF by constructing the altitudes. And the circumcenter of triangle ABC. We also construct the centroid and orthocenter of triangle ABC. Removing some of the lines for the sake of clarity, we want to prove that points H, I and G all lie on the same line. (Math Open Reference)
Theorem: There exists a line that passes through the centroid, circumcenter, and orthocenter of a triangle.
Proof: First there are some constructions that need to be done. We begin by creating a triangle ABC, then finding the midpoints of ead segment, AB, BC, and CA which are D, F, and E respectively. Then we create a smaller triangle DEF. We will find the orthocenter of triangle DEF by constructing the altitudes. And the circumcenter of triangle ABC. We also construct the centroid and orthocenter of triangle ABC. Removing some of the lines for the sake of clarity, we want to prove that points H, I and G all lie on the same line. (Math Open Reference)
Inscribed Quadrilaterals
Theorem: A quadrilateral ABCD can be inscribed in a circle if and only if a pair of opposite angles is supplementary.
Proof: A quadrilateral is inscribed in a circle if and only if the opposite angles are supplementary. (i.e. in quadrilateral ABCD measure angle A+measure angle C = 180 and measure angle B + measure angle D = 180.) This is because the inscribed angle measure is half the measure of the intercepted arc. In particular mArc(BCD)=2 measure angle (A) and mArc(DAB)=2 measure angle(C).
mArc(BCD)+mArc(DAB)=360 degrees.
2 measure angle(A)+2 measure angle(C)=360 degrees, so measure angle(A)+measure angle(C) = 180 degrees, so they are supplementary. (Dwiggins)
Theorem: A quadrilateral ABCD can be inscribed in a circle if and only if a pair of opposite angles is supplementary.
Proof: A quadrilateral is inscribed in a circle if and only if the opposite angles are supplementary. (i.e. in quadrilateral ABCD measure angle A+measure angle C = 180 and measure angle B + measure angle D = 180.) This is because the inscribed angle measure is half the measure of the intercepted arc. In particular mArc(BCD)=2 measure angle (A) and mArc(DAB)=2 measure angle(C).
mArc(BCD)+mArc(DAB)=360 degrees.
2 measure angle(A)+2 measure angle(C)=360 degrees, so measure angle(A)+measure angle(C) = 180 degrees, so they are supplementary. (Dwiggins)
Unique circle for three points
Theorem: For three non colinear points A, B, and C, there is a unique circle that passes through all of three points.
Proof: Connect AB and BC. Draw a perpendicular bisector of AB at the midpoint E as well as a perpendicular bisector of BC at the midpoint F. Since AB and BC are not parallel, their perpendicular bisectors will also not be parallel and must intersect at some point. Let's call that point O. Due to the perpendicular bisector theorem, any point on the perpendicular bisector of AB will be equidistant from points A and B. Therefore, OA congruent to OB. Similarly, for the perpendicular bisector of BC, OB congruent to OC. Then OA congruent to OB congruent to OC. Now we can draw a circle with center O and radius OA that will pass through points B and C since they are all equal distances from center O. O is a unique point because it is the intersection of perpendicular bisectors of AB and BC. Hence, we can say that there is one and only one circle that passes through all three non-colinear points! (Sal Khan, Khan Academy)
Theorem: For three non colinear points A, B, and C, there is a unique circle that passes through all of three points.
Proof: Connect AB and BC. Draw a perpendicular bisector of AB at the midpoint E as well as a perpendicular bisector of BC at the midpoint F. Since AB and BC are not parallel, their perpendicular bisectors will also not be parallel and must intersect at some point. Let's call that point O. Due to the perpendicular bisector theorem, any point on the perpendicular bisector of AB will be equidistant from points A and B. Therefore, OA congruent to OB. Similarly, for the perpendicular bisector of BC, OB congruent to OC. Then OA congruent to OB congruent to OC. Now we can draw a circle with center O and radius OA that will pass through points B and C since they are all equal distances from center O. O is a unique point because it is the intersection of perpendicular bisectors of AB and BC. Hence, we can say that there is one and only one circle that passes through all three non-colinear points! (Sal Khan, Khan Academy)
Parallel line through each triangle vertex
Theorem: A line which bisects two sides of a triangle is parallel to the third.
Proof: Given triangle ABC with the midpoints D and E that are located in its sides BC and AC respectively. Draw the straight line segment ED and copy length ED to make segment DF with the same length.
Connect the points B and F by the straight line segment BF.
The triangles EDC and FDB have the congruent vertical angles EDC and FDB, congruent sides DC and DB as halves of the side BC, and congruent sides ED and FD by construction. Therefore, these triangles are congruent by SAS.
By CPCFC, angle ECD congruent to angle DBF.
There are alternate interior angles formed by the transverse line BC so the straight lines AC parallel BF by the strong alternate interior angle theorem. Also by CPCFC, CE congruent to BF of triangle EDC and triangle FDB. Point E is the midpoint of the side AC so AE congruent EC, and CE congruent BF then BF congruent AE by transitivity. Create diagonal EB.
Thus, we have proved that in the quadrilateral ABFE the two opposite sides BF and AE are parallel and have equal length.
At this point we can refer to the geometry fact:
if a quadrilateral has two opposite sides parallel and of equal length, then two other opposite sides of the quadrilateral are parallel and of equal length too.
It implies that the straight lines AB and EF are parallel.
This is exactly what we wanted to prove. Source.
Theorem: A line which bisects two sides of a triangle is parallel to the third.
Proof: Given triangle ABC with the midpoints D and E that are located in its sides BC and AC respectively. Draw the straight line segment ED and copy length ED to make segment DF with the same length.
Connect the points B and F by the straight line segment BF.
The triangles EDC and FDB have the congruent vertical angles EDC and FDB, congruent sides DC and DB as halves of the side BC, and congruent sides ED and FD by construction. Therefore, these triangles are congruent by SAS.
By CPCFC, angle ECD congruent to angle DBF.
There are alternate interior angles formed by the transverse line BC so the straight lines AC parallel BF by the strong alternate interior angle theorem. Also by CPCFC, CE congruent to BF of triangle EDC and triangle FDB. Point E is the midpoint of the side AC so AE congruent EC, and CE congruent BF then BF congruent AE by transitivity. Create diagonal EB.
Thus, we have proved that in the quadrilateral ABFE the two opposite sides BF and AE are parallel and have equal length.
At this point we can refer to the geometry fact:
if a quadrilateral has two opposite sides parallel and of equal length, then two other opposite sides of the quadrilateral are parallel and of equal length too.
It implies that the straight lines AB and EF are parallel.
This is exactly what we wanted to prove. Source.
Concurrent Altitudes
Theorem: The altitudes of a triangle are concurrent.
Proof: Given a triangle ABC. We draw a line GH through vertex C, parallel to AB; GI through vertex B, parallel to AC; and HI through A, parallel to BC. These lines intersect at G, H, and I to form triangle GHI. Now we have quadrilateral quadrilateral ABCH, where the opposite sides AH and BC are parallel, so we know L(AB)=L(HC) because AB and HC are parallel as well, and there is a constant distance between parallel lines. Similarly from quadrilateral ABCG, we can find that L(AB)=L(CG), so L(HC)=L(CG) and C is the midpoint of HG.
We can use the same logic to find that A is the midpoint of HI and B is the midpoint of GI. (AC = BG from the quadrilateral BGCA, and AC = BI from the quadrilateral IBCA, which gives BG = BI). So the points A, B, and C are the midpoints of the sides of triangle GHI, and the altitudes of triangle ABC are perpendiculars to the sides of triangle GHI at the midpoints of its sides. Therefore, the altitudes are concurrent because perpendicular bisectors of the sides of triangles are concurrent. (Ikleyn, "Lesson Altitudes of a Triangle Are Concurrent.")
Concurrent Perpendicular Bisectors
Theorem: The perpendicular bisectors of a triangle are concurrent.
Proof: For this proof we start with triangle ABC, and we construct the perpendicular bisectors of each side. DG is the perpendicular bisector of AB, EH is that of BC and FI is that of AC. DG and EH are perpendicular to sides AB and BC, respectively, so they cannot be parallel, since AB and BC are not parallel. Since DG and EH are obviously not the same line, they must intersect at one point, P. All points on the perpendicular bisector DG are equidistant to the end points A, B of the segment they bisect; in particular P is equidistant from A and B, therefore AP congruent to BP. By the same logic, we can apply the rule of equidistance to EH and BC, getting the result that BP congruent to CP, which allows us to say CP congruent to AP. This implies that the intersection point P lies on the perpendicular bisector FI of AC. So we have proved that all perpendicular bisectors of triangle ABC pass through point P, and are therefore concurrent. (Ikleyn, "Lesson Perpendicular Bisectors of a Triangle Are Concurrent.")
Theorem: The perpendicular bisectors of a triangle are concurrent.
Proof: For this proof we start with triangle ABC, and we construct the perpendicular bisectors of each side. DG is the perpendicular bisector of AB, EH is that of BC and FI is that of AC. DG and EH are perpendicular to sides AB and BC, respectively, so they cannot be parallel, since AB and BC are not parallel. Since DG and EH are obviously not the same line, they must intersect at one point, P. All points on the perpendicular bisector DG are equidistant to the end points A, B of the segment they bisect; in particular P is equidistant from A and B, therefore AP congruent to BP. By the same logic, we can apply the rule of equidistance to EH and BC, getting the result that BP congruent to CP, which allows us to say CP congruent to AP. This implies that the intersection point P lies on the perpendicular bisector FI of AC. So we have proved that all perpendicular bisectors of triangle ABC pass through point P, and are therefore concurrent. (Ikleyn, "Lesson Perpendicular Bisectors of a Triangle Are Concurrent.")
Midpoint Theorem
Theorem: If D is the midpoint of BC, then AH=2OD.
Proof: If the perpendicular at C to BC intersects the circumcircle again at P’ then BP’ is a diameter of the circumcircle, and so it must pass through the circumcenter O. Since P’AB is also a right angle (by Thales Circle), P’A is parallel to CH. Hence P’AHC is a parallelogram, and AH is equal P’C. Finally, in the right triangle BP’C, OD bisects both BP’ and BC, so we have P’C is equal to 2OD, and so AH is equal to 2OD by transitivity. (Pedoe, "Circles: A Mathematical View)
Diameters of Circle from Diagonals of Rectangles
Theorem: Segments connecting the opposite vertices of two different rectangles form diameters of one circle
Proof: It is important that the two rectangles share two vertices, and so have one concurrent diagonal. So FN is diagonal of both rectangles, DL is diagonal of DNLF and EM is diagonal of MNEF. Diagonals of a rectangle are congruent to each other, so FN congruent to DL, and FN congruent EM, so by transitivity DL congruent EM. Also, diagonals of the same rectangle bisect each other at one point, so FN bisects DL at one point, and FN bisects EM at one point. Because FN is concurrent with itself, the point is the same point. Therefore, FN, EM, and DL can all be bisected by each other to form congruent segments with the same center, forming a circle.
Perpendicular on the Circle:
Theorem from Dan Pedoe: If the two opposite angles of a quadrilateral are right angles, then the four points are concyclic (lying on the same circle. DP perpendicular to LP implies P is on circle with diameter DL. (Pedoe)
Proof: We know angle DLP is a right angle, and we also know we can create a rectangle with side lengths DP and LP, using isometries on the triangle DLP. We already know DL is a diameter of a circle, so if P is concyclic with D, L, it must lie on that same circle.
Theorem from Dan Pedoe: If the two opposite angles of a quadrilateral are right angles, then the four points are concyclic (lying on the same circle. DP perpendicular to LP implies P is on circle with diameter DL. (Pedoe)
Proof: We know angle DLP is a right angle, and we also know we can create a rectangle with side lengths DP and LP, using isometries on the triangle DLP. We already know DL is a diameter of a circle, so if P is concyclic with D, L, it must lie on that same circle.