Question

5. Suppose that the incenter I of ABC is on the triangle’s Euler line. Show that the triangle is isosceles.

6. Suppose that three circles of equal radius pass through a common point P, and denote by A, B, and C the three other points where some two of these circles cross. Show that the unique circle passing through A, B, and C has the same radius as the original three circles.

7. Suppose A, B, and C are distinct points on a line `, and A' , B' , and C/ are distinct points not on ` such that the points D = AB' ∩ A'B, E = AC' ∩ A'C, and F = BC' ∩ B'C exist and are collinear. Show that A' , B' , and C' are also collinear.

8. Use Menelaus’ Theorem to prove Ceva’s Theorem.

9. Join each vertex of ABC to the points dividing the opposite side into equal thirds and let X, Y , and Z be the points of intersection of the pairs of these lines closest to BC, AC, and AB, respectively. Show that the sides of XY Z are parallel to corresponding sides of ABC and that XY Z ∼ ABC.

10. Suppose D, E, and F are points on the sides BC, AC, and AB of ABC, respectively, such that AD, BE, and CF all meet at a point P inside the triangle. Let Q be the point in which AD meets EF. Show that |AQ| · |P D| = |P Q| · |AD|.

11. Suppose that the convex quadrangles ABCD and A0B0C 0D0 are in perspective from the point P, with AA' , BB' , CC' , and DD' all intersecting at P, and suppose that S = AB ∩ A'B' , T = BC ∩ B'C' , U = CD ∩ C'D' , and V = DA ∩ D'A' all exist. Show that S, T, U, and V must all be collinear or give an example showing that this need not be the case.

Answer 1

Q.No.5 :

Proof :

Below is the proof by contradiction.

Let the incenter I lie on the Euler line of triangle ABC.

It is known that orthocenter Hand circumcenter O are isogonal conjugates, i.e. AI is the bisector of angle HAO.

So, (if the point AA does not lie on Euler line) HA/AO=HI/IO (angle bisector theorem). Also HB/BO=HC/CO=HI/IO=HA/AO. And we know that all points X, such that YX/ZX=const, lie on a circle with center on the line YZ (Appolonius circle)

So, A,B,C and I lies on the same circle, and that cannot be true. We have assumed that all of points A,B,C don't lie on the Euler line, so, one of them lies on Euler line and that means ABC is isosceles.

Q.E.D.