# Thread: Geometry of a circle

1. ## Geometry of a circle

1) In a circle, O is the centre of the circle. There are two adjacent chords AB and BC, which connect to form a triangle ABC. Another triangle is also formed off chord AB when it connects to O, forming triangle AOB. If angle OAB = angle BCA, find angle AOB.

2) In a circle, AC and DB are two chords, which intersect at X. AX = CX, so prove AC // DB.

3) Find the 5 unknown angles in these 4 circles.

Thanks for any help.

2. See Soroban's post #5

1) I assume that this is to be set up like a geometric proof, so here goes:

1. $m\angle AOB=m\widehat {AB}$
A central angle is equal to the measure of its intercepted arc.

2. $m\angle BCA=\frac{1}{2}m\widehat{AB}$
An inscribed angle is equal to one-half the measure of its intercepted arc.

3. $m\angle OAB=m\angle BCA$
Given

4. $m\angle OAB=\frac{1}{2}m\widehat{AB}$
Substitution using steps 2 and 3.

5. $m\angle BCA+m\angle OAB=\frac{1}{2}m\widehat{AB}+\frac{1}{2}m\widehat{ AB}$
Addition using steps 2 and 4.

6. $m\angle BCA+m\angle AOB=m\widehat{AB}$

7. $m\angle BCA+m\angle AOB=m\angle AOB$
Substitution using steps 1 and 6

8. $\boxed{m\angle AOB=m\angle BCA + m\angle OAB}$
Symmetric property of equality

3. Originally Posted by BG5965
2) In a circle, AC and DB are two chords, which intersect at X. AX = CX, so prove AC // DB.

1. $m\angle ACD=\frac{1}{2}m\widehat{AD}$
An inscribed angle is equal to one-half of its intercepted arc.

2. $m\angle ABD=\frac{1}{2}m\widehat{AD}$
An inscribed angle is equal to one-half of its intercepted arc.

3. $m\angle ACD=m\angle ABD$
Substitution using steps 1 and 2.

4. $m\angle CAB=\frac{1}{2}m\widehat{BC}$
An inscribed angle is equal to one-half of its intercepted arc.

5. $m\angle CDB=\frac{1}{2}m\widehat{BC}$
An inscribed angle is equal to one-half of its intercepted arc.

6. $m\angle CAB=m\angle CDB$
Substitution using steps 4 and 5.

7. $\triangle ABC$ is isosceles.
Definition of isosceles triangle.

8. $m\angle CAB=m\angle ACD$
Base angles of an isosceles triangle are equal in measure.

9. $m\angle CAB=m\angle ABD$
Substitution using steps 3 and 8.

10. $\overline{AB}\parallel \overline{DB}$
If two lines are cut by a transversal such that alternate interior angles are equal in measue, then the lines are parallel.

4. 3a) Not drawn to scale

$m\widehat{AB}=100$
$m\widehat{BC}=120$
The measure of an inscribed angle is equal to one-half its intercepted arc. So the measure of the intercepted arc is twice the measure of the inscribed angle.

$m\widehat{AC}=360^{\circ}-100^{\circ}-120^{\circ}=140^{\circ}$

$\boxed{x=70}$
The measure of an inscribed angle is equal to one-half its intercepted arc.

5. Hello, BG5965!

1) In a circle, $O$ is the centre of the circle.
There are two adjacent chords $AB$ and $BC$, which connect to form $\Delta ABC.$
Another triangle is also formed off chord $AB$ when it connects to $O$, forming $\Delta AOB.$

If $\angle OAB = \angle BCA$, find $\angle AOB.$

A central angle is measured by its intercepted arc.
. . $\angle AOB \;\;^m_=\;\;\overline{AB}$

An inscribed angle is measured by one-half its intercepted arc.
. . $\angle BCA \;\;^m_=\;\;\tfrac{1}{2}\overline{AB}$

Hence: . $\angle AOB \:=\:2\!\cdot\!\angle BCA$

In $\Delta OAB,\;OA \:=\:OB \quad\Rightarrow\quad \angle OAB \:=\: \angle OBA \:=\:\angle BCA$

$\text{In }\Delta AOB\!:\;\;\underbrace{\angle AOB} + \underbrace{\angle OAB} + \underbrace{\angle OBA} \:=\:180^o$

. . . . . . . $2\!\cdot\!\angle BCA + \angle BCA + \angle BCA \;=\;180^o$

Hence: . $4\!\cdot\!\angle BCA \:=\:180^o \quad\Rightarrow\quad \angle BCA \:=\:45^o$

Therefore: . $\angle AOB \;=\;2\!\cdot\!\angle BCA \;=\;90^o$

6. 3b) Not drawn to scale

$\triangle COB$ is isosceles since radii of the circle are equal (OB=OC)

$m\angle OBC=36^{\circ}$ because base angles of an isosceles triangle are equal in measure.

$m\angle COB=108^{\circ}$. Since there are 180 degrees in a triangle, 180 - 36 - 36 = 108.

$m\widehat{BC}=108^{\circ}$. The measure of a central angle is equal to the measure of its intercepted arc.

$m\angle A=x=\frac{1}{2}m\widehat{BC}=\frac{1}{2}(108^{\cir c})=54^{\circ}$

7. 3c) Angles A and B are right angles because inscribed angles in a semicircle are 90 degrees.

x is complementary to 36 degrees. Therefore x = 54 degrees.

The vertical angle across from x is also 54 degrees because vertical angles have equal measure.

y is complementary to 54 degrees. Therefore y = 36 degrees.

8. 3d) Triangle AOB is isosceles since radii of the same circle are equal in length.

Base angles of an isosceles triangle are equal in measure, so that makes the vertex angle at O = 180 - 23 - 23 = 134 degrees

The measure of arc AB = the measure of the central angle = 134 degrees

c is an inscribed angle.

c = 1/2 the measure of its intercepted arc = 1/2 (134) = 67 degrees.