Question

In a regular heptagon $ABCDEFG$ which of the following is/are true:

• A. $\frac{1}{AB}$ $=$ $\frac{1}{AC}$ $+$ $\frac{1}{AD}$
• B. $\frac{1}{A{B}^2}$ $=$ $\frac{1}{A{C}^2}$ $+$ $\frac{1}{A{D}^2}$
• C. $\frac{1}{AB}$ $=$ $\frac{1}{AC}$ $-$ $\frac{1}{AD}$
• D. $\frac{1}{A{B}^2}$ $=$ $\frac{1}{A{C}^2}$ $-$ $\frac{1}{A{D}^2}$

Answer 1

Answer: (A)

$\frac{1}{AB}$ $=$ $\frac{1}{AC}$ $+$ $\frac{1}{AD}$

Let $ABCDEFG$ be the regular heptagon.

Consider the quadrilateral $ABCD$.

If $a$, $b$ and $c$ represent the lengths of the side, the short diagonal, and the long diagonal respectively, then the lengths of the sides of quadrilateral $ABCE$ are $a$,$a$ ,$b$ and $c$; the diagonals of $ABCE$ are $b$ and $c$, respectively.

Now, Ptolemy's Theorem states that $ab+ac=bc$, which is equivalent to $\frac{1}{a}$ $=$ $\frac{1}{b}$ $+$ $\frac{1}{c}$ upon division by $abc$.

Thus, $\frac{1}{AB}$ $=$ $\frac{1}{AC}$ $+$ $\frac{1}{AD}$ is the correct answer.