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### Area of an Inscribed Quadrilateral

```
Date: 05/11/2000 at 23:06:28
From: Ben
Subject: Geometry

Given: A quadrilateral where the midpoints of each side are connected
to form a new quadrilateral inside the first. What is the ratio of the
area of the larger quadrilateral to the smaller one?

I know that the ratio is 2:1; I solved it by drawing the quadrilateral
as a square and from there the problem was easy. However, I want to
know if there was a way to solve this problem without drawing the

```

```
Date: 05/12/2000 at 02:26:36
From: Doctor Floor
Subject: Re: Geometry

Hi Ben, thanks for writing.

Let us consider a quadrilateral as you described it:

D
/   \
/      M
/         \
N           C
/            |
/             L
/              |
A-------K-------B

First note that, for instance, KL is parallel to and half the measure
of AC, because in triangle ABC segment KL is a midsegment.

The same can be said about MN.

So MN and KL are parallel and congruent. From this we see that KLMN is
a parallelogram.

Now let X and Y be the intersections of AC with KN and LM,
respectively. And let E be the foot of the altitude from B on AC.

KLYX is a parallelogram too. When KL is taken as base, then the
measure of the height of this parallelogram is half BE. We find:

Area KLYX = KL * 0.5*BE
= 0.5*AC * 0.5*BE
= 0.5 * (0.5*AC*BE)
= 0.5 * Area ABC

In the same way we find that area XYMN = 0.5* area ACD.

Combining these two we find the desired result that:

Area KLMN = 0.5 * Area ABCD.

If you need more help, write us back.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 05/14/2000 at 22:39:45
From: Ben
Subject: Geometry

Can you please explain in more depth the reason why when the midpoints
in a 2:1 ratio with the first?

You showed me how the inscribed quadrilateral was a parallelogram.
This was very helpful, but I didn't understand some of the equations
that you wrote for finding the area ratios.

Thank you very much.
```

```
Date: 05/15/2000 at 05:39:13
From: Doctor Floor
Subject: Re: Geometry

Hi again, Ben,

Let me explain the clue of my earlier proof using a triangle:

C
/ | \
G---X---F
/  \  |    \ \
A-----D-Y-----E---B

F and G are the midpoints of BC and AC respectively. DEFG is a
parallelogram; DE and GF have half the lengths of AB. CY is the
altitude from C to AB, so CY and AB are perpendicular. We can use XY
as an altitude in DEFG, and the length of XY is half the length of CY
(in triangle GFC all measures are half the measures in triangle ABC,
so CX is half of CY, and XY is the other half).

So

Area(ABC) = 0.5*AB*CY

and

Area(DEFG) = DE*XY = 0.5*AB * 0.5*CY = 0.5*area(ABC)
|-----v-----|
Area(ABC)

When you apply this twice in the quadrilateral (dividing the
quadrilateral into two triangles by a diagonal) you get the desired
result.

I hope this clears everything up.

Best regards,
- Doctor Floor, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
High School Euclidean/Plane Geometry
High School Geometry
High School Triangles and Other Polygons

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