# Proof

## To Prove:

First we need to create a triangle given 3 points, much like in the construction. We need to find the nine points. After we do that we need to show how the nine point do indeed lie on a the same circle.

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## Proof:

• Draw a line LM and PR.

• We know that L and M bisect the sides AB and BC respectively.

• We also know that if a line bisects one side of a triangle and the other side then that line must be parallel to the 3rd side, therefore we know that LM must be parallel to AC.

• And PR is parallel to AC by a similar argument. Since LM || AC and PR || AC then LM || PR.

• LMRP is a parallelogram

• We also want to show that MR || LP. This can be done by using triangle BEC.

• We know that it is a right angle triangle since BE is an altitude.

• Similar to the above argument M bisects BC therefore if we were to extend MR to touch EC , we know that MR || BE.

• We can have the same argument using triangle ABE and show that LP || BE. We can conclude that LP || MR.

• Now we have a parallelogram with the opposite sides parallel.

• We need to show that LMRP is a rectangle, to do that all angles need to be right angles.

• Since BE is perpendicular to AC, by construction.

• We also know that BC is also perpendicular to PR and LM since all of them are parallel to one another, therefore all the angles in the parallelogram are right angles, making it a rectangle.

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• Construct diagonals in the rectangle and call the intersection point U'. Since LMRP is a rectangle we know that  LR= MP.

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• Just like LMRP was constructed and shown to be a rectangle, QRNL can also be shown the same way to be a rectangle using similar arguments.

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• Also draw diagnoals within this rectangle, but call this intersection point U. We know that QN = LR. Once we have done this we want to show that U=U'.

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• Notice that in both of the rectangles, LMPR and QRNL there is a common diagonal, which is LR.

• So if we overlap both of the rectangles we can see that the line segment LR is shared between both rectangles (highlighted in green). We also know that LR = MP and LR = QN therefore MP = QN.

• So we know that U = U'.

• Since we know that these both are rectangles, we also can know that LU'=U'R=PU'=U'M and we also know that  QU=UN=LU=UR.

• And we know that U=U', so if we substitute we know that LU=UR=PU=UM=QU=UN=LU=UR.

• And if we go back to the properties of circles, these line segments could be considered 6 radii of a circle.

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• Lets focus on the triangle LMR we know that angle M is a right angle since it belongs to the rectangle.

• We also know that in a circle that is there is an inscribed angle of 90 degree, that that chord it extends from must be the diameter, so hence M belongs on the circle.

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• If we take triangle QRM we know that R is also a right angle since it also belongs to the rectangle. And by the same argument above we know that R also belongs to the circle.

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• If we continue this for the remaining triangles, we find that the points L, P, N, R, M, Q, are all points that lie on a circle with centre U.

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• We need to find three more points.

• Let's extend the line from Q to E.

• We know that E is a right angle since we constructed it so that is was a foot of the perpendicular, and if we look at the triangle QEN, we know that QN is the diameter, and if E is 90 degree it must also belong to the circle.

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• If we like above, extend the line from P to D , we know that D is a right angle since it was also constructed as a foot of the perpendicular, and by the same argument D must also belong to the circle.

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• The same happens when we draw a line from R to E, E also belongs on the circle.

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• So as desired we have nine points now: P, L, F, Q, D, M, R, N, E. They all lie on one circle with centre U.

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