# 9.3: Points on a common circle

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Recall that the diameter of a circle is a chord that passes thru the center. If $$[XY]$$ is the diameter of a circle with center $$O$$, then $$\measuredangle XOY = \pi$$. Hence Theorem 9.2.1 implies the following:

## Corollary $$\PageIndex{1}$$

Suppose $$\Gamma$$ is a circle with the diameter $$[AB]$$. A triangle $$ABC$$ has right angle at $$C$$ if and only if $$C \in \Gamma$$.

## Exercise $$\PageIndex{1}$$

Given four points $$A, B, A'$$, and $$B'$$, construct a point $$Z$$ such that both angles $$AZB$$ and $$A'ZB'$$ are right.

Hint

Construct the circles $$\Gamma$$ and $$\Gamma'$$ on the diameters $$[AB]$$ and $$[A'B']$$ respectively. By Corollary $$\PageIndex{1}$$, any point $$Z$$ in the intersection $$\Gamma \cap \Gamma'$$ will do.

## Exercise $$\PageIndex{2}$$

Let $$\triangle ABC$$ be a nondegenerate triangle, $$A'$$ and $$B'$$ be foot points of altitudes from $$A$$ and $$B$$ respectfully. Show that the four points $$A, B, A'$$, and $$B'$$ lie on one circle. What is the center of this circle?

Hint

Note that $$\measuredangle AA'B = \pm \dfrac{\pi}{2}$$ and $$\measuredangle AB'B = \pm \dfrac{\pi}{2}$$. Then apply Corollary $$\PageIndex{2}$$ to $$\square AA'BB'$$.

If $$O$$ is the center of the circle, then $$\measuredangle AOB \equiv 2 \cdot \measuredangle AA'B \equiv \pi$$. That is, $$O$$ is the midpoint of $$[AB]$$.

## Exercise $$\PageIndex{3}$$

Assume a line $$\ell$$, a circle with its center on $$\ell$$ and a point $$P \not\in \ell$$ are given. Make a ruler-only construction of the perpendicular to $$\ell$$ from $$P$$.

Hint

Guess the construction from the diagram. To prove it, apply Theorem 8.2.1 and Corollary $$\PageIndex{1}$$.

## Exercise $$\PageIndex{4}$$

Suppose that lines $$\ell$$, $$m$$ and $$n$$ pass thru a point $$P$$; the lines $$\ell$$ and $$m$$ are tangent to a circle $$\Gamma$$ at $$L$$ and $$M$$; the line $$n$$ intersects $$\Gamma$$ at two points $$X$$ and $$Y$$. Let $$N$$ be the midpoint of $$[XY]$$. Show that the points $$P, L, M$$, and $$N$$ lie on one circle.

We say that a quadrangle $$ABCD$$ is inscribed in circle $$\Gamma$$ if all the points $$A, B, C$$, and $$D$$ lie on $$\Gamma$$.

Hint

Denote by $$O$$ the center of $$\Gamma$$. Use Corollary $$\PageIndex{1}$$ to show that the point lie on the circle with diameter $$[PO]$$.

## Corollary $$\PageIndex{2}$$

A nondegenerate quadrangle $$ABCD$$ is inscribed in a circle if and only if

$$2 \cdot \measuredangle ABC \equiv 2 \cdot \measuredangle ADC.$$

Proof

Since $$\square ABCD$$ is nondegenerate, so is $$\triangle ABC$$. Let $$O$$ and $$\Gamma$$ denote the circulcenter and circumcircle of $$\triangle ABC$$ (they exist by Exercise 8.1.1).

According to Theorem 9.2.1,

$$2 \cdot \measuredangle ABC \equiv \measuredangle AOC$$.

From the same theorem, $$D \ni \Gamma$$ if and only if

$$2 \cdot \measuredangle ADC \equiv \measuredangle AOC,$$

hence the result.

## Exercise $$\PageIndex{5}$$

Let $$\Gamma$$ and $$\Gamma'$$ be two circles that intersect at two distinct point: $$A$$ and $$B$$. Assume $$[XY]$$ and $$[X'Y']$$ are the chords of $$\Gamma$$ and $$\Gamma'$$ respectively, such that $$A$$ lies between $$X$$ and $$X'$$ and $$B$$ lies between $$Y$$ and $$Y'$$. Show that $$(XY) \parallel (X'Y')$$.

Hint

Apply Corollary $$\PageIndex{2}$$ twice for $$\square ABYX$$ and $$\square ABY'X'$$ and use the transversal property (Theorem 7.3.1).

This page titled 9.3: Points on a common circle is shared under a CC BY-SA 4.0 license and was authored, remixed, and/or curated by Anton Petrunin via source content that was edited to the style and standards of the LibreTexts platform.