Question

Complete the following proof.

Prove: The opposite sides of a parallelogram are equal.

(fill in the blanks of the equation in the second picture with the correct number/letter/sign based off the first picture.)

Answer 1

$AD=\sqrt{(\boxed{c}-\boxed{0})^2+(d-\boxed{0})^2}=\sqrt{c^2+\boxed{d}^2}\\\\\\ BC=\sqrt{[(\boxed{b}+c)-\boxed{b}]^2+(\boxed{d}\boxed{-}0)^2}=\sqrt{c^2+\boxed{d}^2}\\\\\\ AB=\sqrt{(\boxed{b}-0)^2+(\boxed{0}-\boxed{0})^2}=\sqrt{b^2+\boxed{0}^2}=\sqrt{b^2}\\\\\\ CD=\sqrt{\big[\boxed{c}-(b+c)\big]^2+(\boxed{d}-\boxed{d})^2}=\sqrt{\boxed{b}^2+0^2}=\sqrt{\boxed{b}^2}$

Answer 2

As  AD = BC , AB = CD , the opposite side of the parallelogram are equal.

What is a Parallelogram ?

Parallelogram is a polygon with four sides , The opposite side of the parallelogram is parallel and equal to each other.

It is given to prove , The opposite sides of a parallelogram are equal.

From the given diagram the measure if the sides can be calculated using the distance formula.

$\rm AD = \sqrt {(c-0)^2+ (d-0)^2)} = \sqrt{c^2+d^2}\\\\BC = \sqrt{ ((b+c)-b)^2 + ( d-0)^2} = \sqrt {c^2+d^2}\\\\AB = \sqrt { (b-)^2+ (0-0)^2} = \sqrt{b^2}\\\\CD = \sqrt{(c-(b+c)^2)+ (d-d)^2}= \sqrt{b^2}$

Therefore AD = BC , AB = CD , the opposite side of the parallelogram are equal.

To know more about Parallelogram

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