# Thread: Geometry of a Circle

1. ## Geometry of a Circle

1) ABCDEF is a hexagon inscribed in a circle. What is x + y + z?

2) In this figure, find the value of angle ABC + angle ADC and figure out if ABCD a cyclic quadrilateral. Therefore, find angle BDC

3a) In this figure, AOB is the diameter. Prove that points OAMP are concylic.
3b) Find two angles equal to angle OPA.

Thanks for any help.

2. Hello, BG5965!

1) $\displaystyle ABCD{E}F$ is a hexagon inscribed in a circle. .What is $\displaystyle x + y + z$ ?
An inscribed angle is measured by one-half its intercepted arc.

We have: .$\displaystyle \begin{array}{ccc}x &=& \frac{1}{2}(BC + CD + DE + EF) \\ \\[-4mm] y &=& \frac{1}{2}(AB + DE + EF + FA) \\ \\[-4mm] z &=& \frac{1}{2}(FA + AB + BC + CD) \end{array}$

Add: .$\displaystyle x+y+z\;=\;\tfrac{1}{2}(2AB + 2BC + 2CD + 2DE + 2EF + 2FA)$

. . . . $\displaystyle x+y+z \;= \;AB + BC + CD + DE + EF + FA$

. . . . $\displaystyle x+y+z \;=\;360^o$

3. Hello again, BG5965!

I have 3(b) . . .

3a) In this figure, $\displaystyle AOB$ is the diameter.
Prove that points OAMP are concylic.

3b) Find two angles equal to $\displaystyle \angle OPA.$

I assume that $\displaystyle O$ is the center of the circle.

$\displaystyle OA = OP$ . . . both are radii.
Hence, $\displaystyle \Delta AOP$ is isosceles.
. . Therefore: .$\displaystyle \angle PAO \:=\:\angle OPA$

$\displaystyle \angle APB = 90^o$ . . . It is inscribed in a semicircle.
Right triangles $\displaystyle APB$ and $\displaystyle MOB$ share angle $\displaystyle PBA$
Then: .$\displaystyle \Delta APB \sim \Delta MOB$

. . Therefore: .$\displaystyle \angle BMO \:=\:\angle PAB \:=\:\angle OPA$