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Mathematics LibreTexts

9.3: Points on a common circle

  • Page ID
    23635
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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 \(\PageIndex{1}\).

    截屏2021-02-18 下午2.34.09.png

    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.

    截屏2021-02-18 下午2.13.51.png

    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).

    截屏2021-02-18 下午2.25.02.png

    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 \((XY) \parallel (X'Y')\).

    截屏2021-02-18 下午2.32.08.png

    Hint

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