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Parallelograms (Posted on 2017-09-09) Difficulty: 4 of 5

  
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    
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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
  

Comments: ( You must be logged in to post comments.)
  Subject Author Date
The TriangleBractals2017-09-21 10:49:12
Some ThoughtsPossible approachbroll2017-09-19 09:56:07
Hints/Tipsre: May I have a hint, please?Bractals2017-09-18 22:38:15
Hints/Tipsre: May I have a hint, please?Bractals2017-09-12 11:39:34
May I have a hint, please?Jer2017-09-11 20:52:59
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