Geometry Area Proof (quadrilateral)

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    Area Geometry Proof
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Discussion Overview

The discussion revolves around proving area relationships in quadrilaterals, specifically focusing on a convex quadrilateral ABCD and its properties when extended to a point E. Participants explore geometric constructions and relationships involving areas of triangles and quadrilaterals, particularly in the context of cyclic quadrilaterals and those with perpendicular diagonals.

Discussion Character

  • Exploratory
  • Technical explanation
  • Conceptual clarification
  • Debate/contested

Main Points Raised

  • One participant describes a construction involving point E on the extension of BC to show that the area of quadrilateral ABCD equals the area of triangle ABE, but struggles with the next steps.
  • Another participant suggests drawing a line through D parallel to AC to find point E, indicating that this construction helps visualize the area relationship between ABE and ABCD.
  • A separate participant presents a problem involving a cyclic quadrilateral ABCD with perpendicular diagonals, aiming to prove that the area of quadrilateral ABCO equals the area of quadrilateral AOCD, but is uncertain about correlating the areas.
  • One participant notes that a quadrilateral with perpendicular diagonals can be classified as a kite, leading to congruent adjacent sides, and discusses the implications of angles formed at the center of the circle.
  • Another participant raises a counterexample of a cyclic quadrilateral that does not have adjacent sides congruent, while also noting the relationship between angles ADC and AOC, referencing Thales' theorem.

Areas of Agreement / Disagreement

Participants express differing views on the properties of cyclic quadrilaterals and the implications of congruent sides. The discussion remains unresolved regarding the area proofs and the validity of the counterexample presented.

Contextual Notes

Participants have not fully resolved the assumptions regarding the properties of cyclic quadrilaterals and the specific conditions under which the area relationships hold. There are also unresolved mathematical steps in the area proofs being discussed.

albertoid
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I'm having trouble with this one:

"Given a convex quadrilateral ABCD, construct a point E on the extension of BC such that the area (ABCD) = (ABE)"

The quadrilateral I drew has A, B, C, D labelled counter-clockwise, and an extension drawn from A to E, where E is BC extended and F is the intersection of AE and DC


I've gotten so far as to figure out i have to construct E so that (ADF) must be equal to the area CEF. I also think that F must be the midpoint of CD, but I'm stuck on what to do now.

Any suggestions or hints would be appreciated :)
 
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Draw a line through D parallel to AC. E is the point where that line intersects BC.

Draw a picture and you should be able to see why the area of ABE is the same as the area of ABCD. (The line AC divides ABCD into two triangles.)
 
Thanks, HallsofIvy :) I had another question.

ABCD is a quadrilateral with perpendicular diagonals, and is inscribed in a circle with center at the point O. Prove that [ABCO] = [AOCD].

I have drawn a cyclic quadrilateral with the vertices AC and BD intersecting at 90 degrees. (Vertices labelled clockwise)

I've broken down the areas of
[ABCO] = [ABO] + [BOC]
[AOCD] = [AOD] + [COD]

However I'm not sure what the next step to take is. I know AO, BO, CO, DO are all radii, but I'm having trouble correlating areas between the two quadrilaterals.
 
It is fairly easy to show that a quadrilateral with diagonals AC and BD intersecting at 90 degrees is a "kite"- that is, it has at least two pairs of congruent adjacent sides. Here, we must have AB= BC and AD= DC. Further, since angle AOC has vertex at the center while while angle ADC has center on the circle, the measure of angle AOC is twice that of angle ADC.
 
but this quadrilateral is inscribed in a circle. I seem to have constructed a counterexample of a quadrilateral (inscribed in a circle) that does not have adjacent sides congruent.

However I did realize that m<ADC is 1/2 m<AOC (Thales: subtended by chord AC). I'm not sure it can be used for this area proof.
 

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