1. Nov 5, 2009

### 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 :)

2. Nov 6, 2009

### 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.)

3. Nov 6, 2009

### 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.

4. Nov 6, 2009

### 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.

5. Nov 7, 2009

### 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.