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The Collinearity of Feet of Perpendiculars to Bisectors - pending

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Also known as: The Midline Property of Angle Bisectors Here is the statement: When you drop perpendiculars from a vertex to the internal and external bisectors of the other two angles, the four feet always lie on a single straight line (the midline of the triangle). Proof: Internal angle bisector of B is the line which sits midway between AB and BC. AB reflected in this will overlap with BC. If you take A's reflection in this which maps to A', A' will lie somewhere on BC. And the internal bisector of B will be the perpendicular bisector of AA'. Let Q be the midpoint of AA'. Similarly A'' is A's reflection in angle B's external bisector. A'' will lie on BC. We can again show that if AA'' has P as its midpoint then BP is the perpendicular bisector AA''. Let B be (0,0). A' = (x1,0), A'' = (x2,0), A = (x1,y1) y co-ordinates of P and Q will be same  = y1/2 Midpoints of AB and AC will also have the same y co-ordinate. So...
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Sides of Orthic triangle DEF are anti parallel to ABC

If D,E,F are the feet of altitudes from A,B,C to BC,CA,AB then sides of DEF are anti parallel to ABC. What does it mean? Let's focus on BCEF quadrilateral which is cyclic. Angle AEF = B and angle AFE = C. If these angles were swapped then EF would be || to BC. But since they are in the exact opposite order it's called anti parallel. Same can be shown for other sides. Also in the triangle DEF, extending the argument above: Angle D = 180 - 2A Angle E = 180 - 2B Angle F = 180 - 2C Assuming that ABC is an acute angle triangle.
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Practice problems pending

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Q1. If the medians BE and CF are equal in a △ABC prove that AB=AC.  S1. Let G be the centroid and let CG = 2x = BG. Let EG = FG = x. Now triangles EGC and FGB are congruent by SAS. Why? Since EG = FG, CG = BG and angle FGB = angle CGE because they are vertical angles. => FB = EC And BF = FA since F is midpoint of AB, similarly AE = EC. => AF + FB = AE + EC => AB = AC H.P. Q2. Let (D, E, F) be the feet of the altitude from (A, B, C) in a (\triangle ABC). Prove that the perpendicular bisector of (EF) also bisects (BC). S2. BCEF is a cyclic quadrilateral with BC as diameter. Perpendicular bisector of EF will pass through the center of the circle since EF is a chord. Center lies on the midpoint of BC as it is a diameter. Hence perpendicular bisector of EF will pass through the midpoint of BC. H.P. Q3. If  S S  is the circumcentre of a  △ A B C △ A BC  and  D , E , F D , E , F  are the feet of the altitudes of  △ A B C △ A BC  th...
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