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 Parallelograms (Posted on 2017-09-09)

Let ABC be an acute triangle with altitudes AD and BE. The
intersection of AD and BE is H. Points F and G make CADF
and CBEG into parallelograms. M is the midpoint of FG.
Ray MC intersects the circumcircle of ΔABC again at point N.

Prove that ANBH is a parallelogram.

 Submitted by Bractals No Rating Solution: (Hide) The points M, C, and O (the circumcenter of ΔABC) are collinear if the distance from M to the line OC is zero. This is true if    |CF|*sin(∠MCF) = |CG|*sin(∠MCG)     (1) From CADF and CBEG (1) becomes    |AD|*sin(∠MCF) = |BE|*sin(∠MCG)     (2) At point C we see that ∠MCF + ∠OCB = 90° and ∠MCG + ∠OCA = 90°. Thus,    sin(∠MCF) = cos(∠OCB) and sin(∠MCG) = cos(∠OCA) Equation (2) then becomes    |AD|*cos(∠OCB) = |BE|*cos(∠OCA)     (3) From ΔADC and ΔBEC:    |AD| = |AC|*sin(∠ACD) and |BE| = |BC|*sin(∠BCE) Since ∠ACD = ∠BCE, (3) becomes    |AC|*cos(∠OCB) = |BC|*cos(∠OCA) This in turn becomes,    |AC|*(|BC|/(2*|OC|)) = |BC|*(|AC|/(2*|OC|)). Therefore, Points M, C, and O are collinear. Since CN contains point O, it is a diameter of the circumcircle. Therefore, ∠CAN and ∠CBN are right angles since they are subtended by CN.    AN and BE perpendicular to AC ⇒ AN || BE ⇒ AN || BH    BN and AD perpendicular to BC ⇒ NB || DA ⇒ NB || HA Therefore, ANBH is a parallelogram. QED

 Subject Author Date The Triangle Bractals 2017-09-21 10:49:12 Possible approach broll 2017-09-19 09:56:07 re: May I have a hint, please? Bractals 2017-09-18 22:38:15 re: May I have a hint, please? Bractals 2017-09-12 11:39:34 May I have a hint, please? Jer 2017-09-11 20:52:59

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