Prove it if you can.?

Question

prove that area of a quadrilateral formed by joining mid point of any quadrilateral is half of the original quadrilateral.

Answer 1

Not too hard. The question is, given a quadrilateral ABCD with midpoints EFGH (E is between A and B), show the area of EFGH is 1/2 the area of ABCD.

First draw AC. Triangle ABC is similar to EBF, and each side is twice as long, so the area of EBF = 1/4 area of ABC. Do the same thing for ADC and HDG to get EBF + HDG = 1/4(ABC + ADC). But ABC + ADC is the area of the entire quadrilateral, so the area of the 2 triangles is 1/4 the area of the quadrilateral.

Now draw BD and do the same thing to show FCG + HAE = 1/4(BCD + DAB) = 1/4 ABCD. Adding these 2 gives the area of the 4 triangles is 1/2 the area of the quadrilateral. The midpoint quadrilateral is the qhole quadrilateral minus this, or 1/2 the area of the quadrilateral.

Answer 2

Trapezoids and parallelograms are easy:
For a trapezoid or parallelogram abcd, b & d parallel,
A = h(b + d)/2, where h is the altitude, or perpendicular distance between b & d. Joining consecutive midpoints yields 4 triangles as well as the new quadrilateral. The sum of the areas of the triangles is:
(1/2)(b/2)(h/2) + (1/2)(b/2)(h/2) +
(1/2)(d/2)(h/2) + (1/2)(d/2)(h/2) =
(1/4)hb + (1/4)hd = h(b + d)/4, which is half the area of the parent figure, so the area of the inscribed quadrilateral is also half that of the parent figure.

The question remains: Can this be extended to the general convex quadrilateral? (It's rather easy to show that it does not apply to concave or crossed quadrilaterals.) You can do an inverse inductive proof by constructing a superscribing trapeziod, but that still does not prove the general case.

If I can come up with a general proof before you terminate the question, I'll edit this.

Edit:
Unh, duh! The proof was in the answer above by sofarsogood, but I just couldn't see it! Triangle AEH has 1/4 the area of triangle ABD. Triangle BFE has 1/4 the area of triangle BCA. Triangle CGF has 1/4 the area of triangle CDB. Triangle DHG has 1/4 the area of triangle DAC. Adding the areas of the 4 small triangles yields 1/4 the area of the 4 large triangles, but the area of the 4 large triagles is twice the area of the quadrilateral, so the 4 small triangles remove half this area leaving the area of 1/2 the parent figure for the inscribed figure.

Answer 3

yes...this is true...when the quadrilateral you are talking about
is formed into two equal triangles meeting at the midpoint that's making the end of one side of each triangle....Talk about mirror image of a triangle...

i hope you got the idea...

Answer 4

Unsolvable question: join the mid-point of the quadrilateral to what? Or do you mean join the mid-pointS (plural)? and even then...mid-points of two opposing sides? two adjacent sides?

Answer 5

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