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December 28th, 2017, 06:24 PM   #1
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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
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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.
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Last edited by skipjack; December 29th, 2017 at 10:19 AM.
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December 29th, 2017, 09:54 AM   #3
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https://en.wikipedia.org/wiki/Brahmagupta_theorem
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December 29th, 2017, 07:11 PM   #4
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Good morning !
Quote:
Originally Posted by skipjack View Post
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 !
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December 29th, 2017, 07:16 PM   #5
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Salam !

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Originally Posted by greg1313 View Post
Thanks for this useful information. I didn't expected that it is already a known theorem.
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