Question

The area of a parallel is given by the formula A = bh, where A is the area, b is the length of a base, and h is the height perpendicular to the base. ABCD is a parallelogram. E, F, G, H are the midpoints of the sides. Show that the area of EFGH is half the area of ABCD.

Answer 1

Let's begin by listing out the information given to us:

$\begin{gathered} A=bh \\ where\colon A=Area,b=base,h=height \\ \\ E=\frac{1}{2}AB,F=\frac{1}{2}BC,G=\frac{1}{2}CD,H=\frac{1}{2}DA \\ Prove\colon Area(EFGH)=\frac{1}{2}\cdot Area(ABCD) \end{gathered}$

We join H & F

Line AD is parallel to Line BC (the opposite sides of a parallelogram are congruent);

$\begin{gathered} AD=BC\Rightarrow\frac{1}{2}AD=\frac{1}{2}BC \\ H\Rightarrow\frac{1}{2}BC \end{gathered}$

Line DH & CF are parallel & congruent:

$DH=CF$

The pair of opposite sides are congruent & parallel. This implies that both DHFC & HABF are parallelograms

Observe that Triangle HGF has the same base as Parallelogram DHFC and both are between the same parallel lines. The area of a triangle is half of a parallelogram if they both share the same base & are parallel.

$\Delta HGF=\frac{1}{2}\cdot Area(DHFC)----1$

In like manner, observe that Triangle HEF has the same base as Parallelogram HABF and both are between the same parallel lines

$\Delta HEF=\frac{1}{2}\cdot Area(HFAB)----2$

We proceed to add equations 1 & 2, we have:

$\begin{gathered} \Delta HGF+_{}\Delta HEF=\frac{1}{2}\cdot Area(HFAB)+\frac{1}{2}\cdot Area(DHFC) \\ Area(HEFG)=\frac{1}{2}\lbrack Area(HFAB)+Area(DHFC)\rbrack \\ Area(HFAB)+Area(DHFC)=Area(ABCD) \\ \\ \therefore Area(HEFG)=\frac{1}{2}\cdot Area(ABCD) \end{gathered}$