My Math Forum Geometry Problem Series, Question 3:

 Geometry Geometry Math Forum

 November 1st, 2019, 08:53 AM #1 Senior Member     Joined: Mar 2015 From: Universe 2.71828i3.14159 Posts: 169 Thanks: 65 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 $BCH$. --------
 November 1st, 2019, 09:11 AM #2 Senior Member   Joined: Jun 2019 From: USA Posts: 386 Thanks: 211 Do you mean AGFH?
November 1st, 2019, 10:00 AM   #3
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Math Focus: Area of Circle
Quote:
 Originally Posted by DarnItJimImAnEngineer Do you mean AGFH?
Quadrilateral $AGFH$.

Triangles $DEG$ and $BCH$.

 November 1st, 2019, 10:11 AM #4 Senior Member   Joined: Jun 2019 From: USA Posts: 386 Thanks: 211 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. )
November 1st, 2019, 12:38 PM   #5
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Math Focus: Area of Circle
Quote:
 Originally Posted by DarnItJimImAnEngineer 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...

 November 1st, 2019, 10:34 PM #6 Global Moderator   Joined: Dec 2006 Posts: 21,124 Thanks: 2332 $\small\triangle$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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