
# 9.2: Inscribed angle


We say that a triangle is inscribed in the circle $$\Gamma$$ if all its vertices lie on $$\Gamma$$.

Theorem $$\PageIndex{1}$$

Let $$\Gamma$$ be a circle with the center $$O$$, and $$X, Y$$ be two distinct points on $$\Gamma$$. Then $$\triangle XPY$$ is inscribed in $$\Gamma$$ if and only if

$2 \cdot \measuredangle XPY \equiv \measuredangle XOY.$

Equivalently, if and only if

$$\measuredangle XPY \equiv \dfrac{1}{2} \cdot \measuredangle XOY$$     or     $$\measuredangle XPY \equiv \pi + \dfrac{1}{2} \cdot \measuredangle XOY.$$

Proof

the "only if" part. Let $$(PQ)$$ be the tangent line to $$\Gamma$$ at $$P$$. By Theorem 9.1.1,

$$2 \cdot \measuredangle QPX \equiv \measuredangle POX$$,   $$2 \cdot \measuredangle QPY \equiv \measuredangle POY.$$

Subtracting one identity from the other, we get 9.2.1.

"If" part. Assume that 9.2.1 holds for some $$P \not\in \Gamma$$. Note that $$\measuredangle XOY \ne 0$$. Therefore, $$\measuredangle XPY \ne 0$$ nor $$\pi$$; that is, $$\measuredangle PXY$$ is nondegenerate.

The line $$(PX)$$ might be tangent to $$\Gamma$$ at the point $$X$$ or intersect $$\Gamma$$ at another point; in the latter case, suppose that $$P'$$ denotes this point of intersection.

In the first case, by Theorem 9.1, we have

$$2 \cdot \measuredangle PXY \equiv \measuredangle XOY \equiv 2 \cdot \measuredangle XPY.$$

Applying the transversal property (Theorem 7.3.1), we get that $$(XY) \parallel (PY)$$, which is impossible since $$\triangle PXY$$ is nondegenerate.

In the second case, applying the "if" part and that $$P, X$$, and $$P'$$ lie on one line (see Exercise 2.4.2) we get that

$$\begin{array} {rcl} {2 \cdot \measuredangle P'PY} & \equiv & {2 \cdot \measuredangle XPY \equiv \measuredangle XOY \equiv} \\ {} & \equiv & {2 \cdot \measuredangle XP'Y \equiv 2 \cdot \measuredangle XP'P.} \end{array}$$

Again, by transversal property, $$(PY) \parallel (P'Y)$$, which is impossible since $$\triangle PXY$$ is nondegenerate.

Exercise $$\PageIndex{1}$$

Let $$X, X', Y$$, and $$Y'$$ be distinct points on the circle $$\Gamma$$. Assume $$(XX')$$ meets $$(YY')$$ at a point $$P$$. Show that

(a) $$2 \cdot \measuredangle XPY \equiv \measuredangle XOY + \measuredangle X'OY'$$;

(b) $$\triangle PXY \sim \triangle PY'X'$$;

(c) $$PX \cdot PX' = |OP^2 - r^2|$$, where $$O$$ is the center and $$r$$ is the radius of $$\Gamma$$.

(The value $$OP^2 - r^2$$ is called the power of the point $$P$$ with respect to the circle $$\Gamma$$. Part (c) of the exercise makes it a useful tool to study circles, but we are not going to consider it further in the book.)

Hint

(a) Apply Theorem $$\PageIndex{1}$$ for $$\angle XX'Y$$ and $$\angle X'YY'$$ and Theorem 7.4.1 for $$\triangle PYX'$$.

(b) If $$P$$ is inside of $$\Gamma$$ then $$P$$ lies between $$X$$ and $$X'$$ and between $$Y$$ and $$Y'$$ in this case $$\angle XPY$$ is vertical to $$\angle X'PY'$$. If $$P$$ is outside of $$\Gamma$$ then $$[PX) = [PX')$$ and $$[PY) = [PY')$$. In both cases we have that $$\measuredangle XPY = \measuredangle X'PY'$$.

Applying Theorem $$\PageIndex{1}$$ and Exercise 2.4.2, we get that

$$2 \cdot \measuredangle Y'X'P \equiv 2 \cdot \measuredangle Y'X'X \equiv 2 \cdot \measuredangle Y'YX \equiv 2 \dot \measuredangle PYX.$$

According to Theorem 3.3.1, $$\angle Y'X'P$$ and $$\angle PYX$$ have the same sign; therefore $$\measuredangle Y'X'P = \measuredangle PYX$$. It remains to apply the AA similarity condition.

(c) Apply (b) assuming $$[YY']$$ is the diameter of $$\Gamma$$.

Exercise $$\PageIndex{2}$$

Three chords $$[XX']$$, $$[YY']$$, and $$[ZZ']$$ of the circle $$\Gamma$$ intersect at a point $$P$$. Show that

$$XY' \cdot ZX' \cdot YZ' = X'Y \cdot Z'X \cdot Y'Z.$$

Hint

Apply Exerciese $$\PageIndex{1} b three times. Exercise \(\PageIndex{3}$$

Let $$\Gamma$$ be a circumcircle of an acute triangle $$ABC$$. Let $$A'$$ and $$B'$$ denote the second points of intersection of the altitudes from $$A$$ and $$B$$ with $$\Gamma$$. Show that $$\triangle A'B'C$$ is isosceles.

Hint

Let $$X$$ and $$Y$$ be the foot points of the altitudes from $$A$$ and $$B$$. Suppose that $$O$$ denotes the circumcenter.

By AA condition, $$\triangle AXC \sim \triangle BYC$$. Then

$$\measuredangle A'OC \equiv 2 \cdot \measuredangle A'AC \equiv - 2 \cdot \measuredangle B'BC \equiv - \measuredangle B'OC.$$

By SAS, $$\triangle A'OC \cong \triangle B'OC$$. Therefore, $$A'C = B'C$$.

Exercise $$\PageIndex{4}$$

Let $$[XY]$$ and $$[X'Y']$$ be two parallel chords of a circle. Show that $$XX' = YY'$$.

Exercise $$\PageIndex{5}$$

Watch “Why is pi here? And why is it squared? A geo- metric answer to the Basel problem” by Grant Sanderson. (It is available on YouTube.)

Prepare one question.