### Video Transcript

Prove that a parallelogram
circumscribing a circle is a rhombus.

The word “circumscribe” comes from
the Latin “circum” meaning “around” and “scribe” meaning “to write.” This means that a parallelogram
circumscribes a circle if the circle within it touches all four sides of the polygon
as shown.

We already know since it’s a
parallelogram that the opposite sides are parallel and of equal length. 𝐴𝐷 and 𝐵𝐶 are parallel and
equal, and 𝐴𝐵 and 𝐶𝐷 are also parallel and equal. To prove that 𝐴𝐵𝐶𝐷 is actually
a rhombus, we need to show that the length of all the sides are equal.

Let’s recall some circle
theorems. We know that the tangents drawn to
a circle from a point are of equal length. Since the parallelogram
circumscribes the circle, that means that all four sides are a tangent to the
circle. And we can add the points where the
tangents meet the circle as 𝐸, 𝐹, 𝐺, and 𝐻.

The line segments joining 𝐴 and 𝐸
and 𝐴 and 𝐹 must therefore be of equal length. The line segments joining 𝐵 to 𝐹
and 𝐵 to 𝐺 are also of equal length. 𝐶𝐺 must be equal to 𝐶𝐻, and
𝐷𝐸 must be also equal to 𝐷𝐻. We can add these four
equations. And in doing so, we can see that
𝐴𝐸 plus 𝐵𝐺 plus 𝐶𝐺 plus 𝐷𝐸 is equal to 𝐴𝐹 plus 𝐵𝐹 plus 𝐶𝐻 plus
𝐷𝐻.

Now the line segments joining 𝐵 to
𝐺 and 𝐶 to 𝐺 all form the line 𝐵𝐶. And the line segments joining 𝐴 to
𝐸 and 𝐷 to 𝐸 form the line 𝐴𝐷. The line segments joining 𝐴 to 𝐹
and 𝐵 to 𝐹 form the line 𝐴𝐵, and the line segments 𝐶 to 𝐻 and 𝐷 to 𝐻 form
the line 𝐶𝐷. So we can rewrite this equation as
𝐵𝐶 plus 𝐴𝐷 equals 𝐴𝐵 plus 𝐶𝐷.

Now since we said that 𝐴𝐷 was
equal to 𝐵𝐶 and 𝐴𝐵 was equal to 𝐶𝐷, we can rewrite this as two individual
equations. We can say two 𝐴𝐷 is equal to two
𝐴𝐵 and two 𝐵𝐶 is equal to two 𝐶𝐷. Dividing each side of our equations
by two, and we get that 𝐴𝐷 is equal to 𝐴𝐵 and 𝐵𝐶 is equal to 𝐶𝐷. We knew 𝐴𝐵𝐶𝐷 was a
parallelogram. However, we have now shown that all
four sides are of equal length. So in fact, 𝐴𝐵𝐶𝐷 is a rhombus
as required.