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November 1st, 2019, 08:53 AM   #1
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Math Focus: Area of Circle
Geometry Problem Series, Question 3:

Pentagon $ABCDE$ is inscribed in a circle.
$AB \parallel EC$, $AE \parallel BD$.
$AD \cap EC \equiv G$, $BD \cap EC \equiv F$ and $AC \cap BD \equiv H$.
Prove that the area of $AGFH$ is equal to the sum of the areas of $DEG$ and $BCH$.
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November 1st, 2019, 09:11 AM   #2
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Do you mean AGFH?
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November 1st, 2019, 10:00 AM   #3
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Math Focus: Area of Circle
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Originally Posted by DarnItJimImAnEngineer View Post
Do you mean AGFH?
Quadrilateral $AGFH$.

Triangles $DEG$ and $BCH$.
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November 1st, 2019, 10:11 AM   #4
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Yeah, I think it said AFGH before you fixed it, unless I was just having a dyslexic moment.

On a side note, if we solve Question 1, 2, and 3, do we unlock the secrets of alchemy?
(Couldn't resist. )
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November 1st, 2019, 12:38 PM   #5
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Math Focus: Area of Circle
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Originally Posted by DarnItJimImAnEngineer View Post
Yeah, I think it said AFGH before you fixed it unless I was just having a dyslexic moment.

On a side note, if we solve Questions 1, 2, and 3, do we unlock the secrets of alchemy?
(Couldn't resist. )
No, but you will unlock the secret of the geometry of the Multiverse...
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November 1st, 2019, 10:34 PM   #6
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$\small\triangle$BFG = $\small\triangle$AFG = $\small\triangle$ADF - $\small\triangle$DFG = $\small\triangle$DEF - $\small\triangle$DFG = $\small\triangle$DEG
Area(AGFH) = $\small\triangle$ACG - $\small\triangle$CFH = $\small\triangle$BCG - $\small\triangle$CFH = $\small\triangle$BFG + $\small\triangle$BCH = $\small\triangle$DEG + $\small\triangle$BCH
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