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| Parallelograms (Posted on 2017-09-09) |
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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.
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Submitted by Bractals
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Solution:
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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
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