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.
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
2⋅∡XPY≡∡XOY.
Equivalently, if and only if
∡XPY≡12⋅∡XOY or ∡XPY≡π+12⋅∡XOY.
- Proof
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the "only if" part. Let (PQ) be the tangent line to Γ at P. By Theorem 9.1.1,
2⋅∡QPX≡∡POX, 2⋅∡QPY≡∡POY.
Subtracting one identity from the other, we get 9.2.1.
"If" part. Assume that 9.2.1 holds for some P∉Γ. Note that ∡XOY≠0. Therefore, ∡XPY≠0 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
2⋅∡PXY≡∡XOY≡2⋅∡XPY.
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
2⋅∡P′PY≡2⋅∡XPY≡∡XOY≡≡2⋅∡XP′Y≡2⋅∡XP′P.
Again, by transversal property, (PY)∥(P′Y), which is impossible since △PXY is nondegenerate.
Let X,X′,Y, and Y′ be distinct points on the circle Γ. Assume (XX′) meets (YY′) at a point P. Show that
(a) 2⋅∡XPY≡∡XOY+∡X′OY′;
(b) △PXY∼△PY′X′;
(c) PX⋅PX′=|OP2−r2|, where O is the center and r is the radius of Γ.
(The value OP2−r2 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
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(a) Apply Theorem 9.2.1 for ∠XX′Y and ∠X′YY′ 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 ∠X′PY′. If P is outside of Γ then [PX)=[PX′) and [PY)=[PY′). In both cases we have that ∡XPY=∡X′PY′.
Applying Theorem 9.2.1 and Exercise 2.4.2, we get that
2⋅∡Y′X′P≡2⋅∡Y′X′X≡2⋅∡Y′YX≡2˙∡PYX.
According to Theorem 3.3.1, ∠Y′X′P and ∠PYX have the same sign; therefore ∡Y′X′P=∡PYX. It remains to apply the AA similarity condition.
(c) Apply (b) assuming [YY′] is the diameter of Γ.
Three chords [XX′], [YY′], and [ZZ′] of the circle Γ intersect at a point P. Show that
XY′⋅ZX′⋅YZ′=X′Y⋅Z′X⋅Y′Z.
- Hint
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Apply Exerciese \(\PageIndex{1} b three times.
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 △A′B′C is isosceles.
- Hint
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Let X and Y be the foot points of the altitudes from A and B. Suppose that O denotes the circumcenter.
By AA condition, △AXC∼△BYC. Then
∡A′OC≡2⋅∡A′AC≡−2⋅∡B′BC≡−∡B′OC.
By SAS, △A′OC≅△B′OC. Therefore, A′C=B′C.
Let [XY] and [X′Y′] be two parallel chords of a circle. Show that XX′=YY′.
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.