parallelogram conjecture?

[BMAD]

Start with any general quadrilateral, and connect the midpoints of consecutive sides, making an inscribed quadrilateral as in the diagram. That inscribed quadrilateral, in the diagram, seems to be a parallelogram. Let me conjecture that this inscribed quadrilateral is a parallelogram with half the area of the original quadrilateral. Can you prove or disprove either part of my conjecture?

[ShadowAngel7]

Which diagram? I'm not seeing one.

[BMAD]

Which diagram? I'm not seeing one.

oops. nice catch. I forgot to upload it. but it is meant to just show an arbitrary quadrilateral, see below:

[bonanova]

Your conjecture is true.

Call the original quadrilateral ABCD.

Draw diagonal AC. The line connecting midpoints of sides AB and BC is parallel to AC (similar triangles.)

Likewise the line connecting midpoints of sides CD and DA is parallel to AC.
Repeat for diagonal BD.

The inscribed quadrilateral comprises two sets of parallel lines and is thus a parallelogram.

[BMAD]

What about the other part of the conjecture that the area is half?

Your conjecture is true.

Call the original quadrilateral ABCD.

Draw diagonal AC. The line connecting midpoints of sides AB and BC is parallel to AC (similar triangles.)

Likewise the line connecting midpoints of sides CD and DA is parallel to AC.

Repeat for diagonal BD.

The inscribed quadrilateral comprises two sets of parallel lines and is thus a parallelogram.

[bonanova]

What about the other part of the conjecture that the area is half?

Your conjecture is true.

Call the original quadrilateral ABCD.

Draw diagonal AC. The line connecting midpoints of sides AB and BC is parallel to AC (similar triangles.)

Likewise the line connecting midpoints of sides CD and DA is parallel to AC.

Repeat for diagonal BD.

The inscribed quadrilateral comprises two sets of parallel lines and is thus a parallelogram.

Consider the triangles ABD and AQP.

The latter triangle has half the base and half the height and thus 1/4 the area of the former.

The same can be said of triangles CDB and CSR.

ABD and CDB sum to the total area.

Thus AQP and CSR sum to 1/4 of the total area, as do DPS and BRQ.

The inscribed quadrilateral comprises the remaining 1/2.