Geometry Area Proof (quadrilateral)

[albertoid]

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 :)

 

[HallsofIvy]

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.)

 

[albertoid]

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.

 

[HallsofIvy]

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.

 

[albertoid]

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.