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December 28th, 2017, 06:24 PM   #1
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Joined: Aug 2011
From: Nouakchott, Mauritania

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Interesting Circle Question

Salam !

This is your Christmas gift from me .

Let $\displaystyle O$ be the center of a circle. $\displaystyle A$, $\displaystyle B$, $\displaystyle C$ and $\displaystyle D$ are points of this circle such that the straight lines $\displaystyle (AC)$ and $\displaystyle (BD)$ are perpendicular and secant in $\displaystyle I$.

Let $\displaystyle H$ be the midpoint of the segment $\displaystyle [BC]$.

In the triangle $\displaystyle IAD$, let $\displaystyle J$ be the foot of the altitude from $\displaystyle I$.

Show that the points $\displaystyle I$, $\displaystyle J$ and $\displaystyle H$ are aligned (collinear).

Good luck !
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 December 29th, 2017, 09:29 AM #2 Global Moderator   Joined: Dec 2006 Posts: 18,408 Thanks: 1460 As HB = HI, $\small\angle$HIB = $\small\angle$HBI = $\small\angle$CAD = $\small\angle$DIJ, and so the points H, I and J are collinear. Thanks from Ould Youbba and johng40 Last edited by skipjack; December 29th, 2017 at 10:19 AM.
 December 29th, 2017, 09:54 AM #3 Global Moderator     Joined: Oct 2008 From: London, Ontario, Canada - The Forest City Posts: 7,691 Thanks: 976 Math Focus: Elementary mathematics and beyond Thanks from Ould Youbba
December 29th, 2017, 07:11 PM   #4
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Joined: Aug 2011
From: Nouakchott, Mauritania

Posts: 81
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Math Focus: Algebra, Cryptography
Good morning !
Quote:
 Originally Posted by skipjack As HB = HI, $\small\angle$HIB = $\small\angle$HBI = $\small\angle$CAD = $\small\angle$DIJ, and so the points H, I and J are collinear.
I like this fast solution ! Well done !

December 29th, 2017, 07:16 PM   #5
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Joined: Aug 2011
From: Nouakchott, Mauritania

Posts: 81
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Math Focus: Algebra, Cryptography
Salam !

Quote:
 Originally Posted by greg1313 https://en.wikipedia.org/wiki/Brahmagupta_theorem
Thanks for this useful information. I didn't expected that it is already a known theorem.

 Tags circle, interesting, question

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