**Feuerbach’s circle: **

**Let ABC be a triangle. Let D,E,F be the midpoints of AC,BC,BA respectively. Let G,H,I be the feet of each altitude in the triangle. Let J be the point where the altitudes meet. Denote the points M,L,K as the midpoints of JA,JC,JB. Then all points D,E,F,G,H,I,K,L,M lie on a circle. **

Proof:

Due to the intercept theorem we can say, that ED is parallel to AB and therefore GF, EF is parallel to AC (DH),and DF is parallel to BC (EI). Let’s prove, that EDGF is isosceles. Since CGA is 90 degrees and D is half of AC, by the Thales theorem (which says, that D is the midpoint of the circle around ACG) DG is equal to AD. By the intercept theorem FE is also half of AC. Because the diagonals have the same length DGFE is an isosceles trapezoid and its vertices lie on a circle. It is possible to use the same argument on DHEF and DFEI. Since D,E,F are in each trapezoid, D,E,F,G,H,I are on a circle. Now we use this information on the triangle BCJ. We proved, that E,I,L,K are on one circle, since E,K,L are the midpoints of the sides of the triangle BCJ and I is the foot of the altitude from J. We can use the same arguments for triangles BJA and CAJ. This completes the proof.

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