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=(1)/(2)AB,F=(1)/(2)BC,G=(1)/(2)CD,H=(1)/(2)DA \\ Prove\colon Area(EFGH)=(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(1)/(2)AD=(1)/(2)BC \\ H\Rightarrow(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=(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=(1)/(2)\cdot Area(HFAB)----2`

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

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