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Mathematics LibreTexts

9.2: Inscribed angle

( \newcommand{\kernel}{\mathrm{null}\,}\)

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

Theorem 9.2.1

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

2XPYXOY.

Equivalently, if and only if

XPY12XOY or XPYπ+12XOY.

Proof

截屏2021-02-18 上午11.19.51.png截屏2021-02-18 上午11.20.15.png截屏2021-02-18 上午11.20.24.png

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

2QPXPOX, 2QPYPOY.

Subtracting one identity from the other, we get 9.2.1.

"If" part. Assume that 9.2.1 holds for some PΓ. Note that XOY0. Therefore, XPY0 nor π; that is, PXY is nondegenerate.

The line (PX) might be tangent to Γ at the point X or intersect Γ 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

2PXYXOY2XPY.

Applying the transversal property (Theorem 7.3.1), we get that (XY)(PY), which is impossible since 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

2PPY2XPYXOY2XPY2XPP.

Again, by transversal property, (PY)(PY), which is impossible since PXY is nondegenerate.

Exercise 9.2.1

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

(a) 2XPYXOY+XOY;

(b) PXYPYX;

(c) PXPX=|OP2r2|, where O is the center and r is the radius of Γ.

截屏2021-02-18 上午11.32.12.png

(The value OP2r2 is called the power of the point P with respect to the circle Γ. 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 9.2.1 for XXY and XYY and Theorem 7.4.1 for PYX.

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

Applying Theorem 9.2.1 and Exercise 2.4.2, we get that

2YXP2YXX2YYX2˙PYX.

According to Theorem 3.3.1, YXP and PYX have the same sign; therefore YXP=PYX. It remains to apply the AA similarity condition.

(c) Apply (b) assuming [YY] is the diameter of Γ.

Exercise 9.2.2

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

XYZXYZ=XYZXYZ.

截屏2021-02-18 下午1.35.07.png

Hint

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

Exercise 9.2.3

Let Γ 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 Γ. Show that ABC is isosceles.

截屏2021-02-18 下午1.36.50.png

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, AXCBYC. Then

AOC2AAC2BBCBOC.

By SAS, AOCBOC. Therefore, AC=BC.

Exercise 9.2.4

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

Exercise 9.2.5

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.


This page titled 9.2: Inscribed angle 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.

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