Tag Archives: inscribed angle theorem

The Proof of the Cyclic Quadrilateral Theorem

Posted on by
Reply

A cyclic quadrilateral is a quadrilateral inscribed in a circle. A polygon that is inscribed in a circle is a polygon whose vertices are on the circle. Some of the inscribed polygons are shown in the next figure.

Examples of polygons inscribed in a circle.

In this post, we are going to show a special property of one inscribed polygon which is the cyclic quadrilateral theorem about angles. We are going to prove that its opposite angles add up to $latex 180 ^\circ$

Theorem

The sum of the measures of the opposite angle of a cyclic quadrilateral is $latex 180 ^\circ$  Continue reading →

Proof That the Angle Sum of a Pentagram is 180 Degrees

Posted on by
2

A pentagram is a five pointed star as shown below. In this post, we are going to learn two proofs that the angle sum of a pentagram is equal to 180 degrees.  In the discussion below, we will refer to the angles at the 5 angles at the tip of the pentagram.

Proof 1: Inscribing the Star in a Circle

If we inscribed the pentagram in a circle, then each of the 5 angles at the vertices is an inscribed angle. Each of the inscribed angles intercepts an arc as shown by the colors in the figure below. The “red angle” intercepts the red arc.

Angle Sum of a Pentagram

Now, notice that the five angles intercept 5 arcs which when combined is the circle. Since the measure of the all the arcs of the circle is 360 degrees, by the inscribed angle theorem, the total angles is half of that which is 180 degrees. Therefore, the angle sum of the interior angles of a pentagram is 180. Continue reading →

The Proof of the Third Case of the Inscribed Angle Theorem

Posted on by
1

This is the third and the last case of the Proofs of the Inscribed Angle Theorem. In the first case, we used an auxiliary line between the sides of the angles to come up with the proof. In the second case, we drew a line segment from the center to the intersection of the circle and one of the sides to form an isosceles triangle. In this case, we cannot do both.

inscribed angle

To prove the third case, we will draw an auxiliary line outside the circle.  We draw segment $latex \overline{BD}$ through the center of the circle. The detail of the proof is as follows. Continue reading →