9.4: Method of additional circle
Assume that two chords \([AA']\) and \([BB']\) intersect at the point \(P\) inside their circle. Let \(X\) be a point such that both angles \(XAA'\) and \(XBB'\) are right. Show that \((XP) \perp (A'B')\).
- Answer
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Set \(Y = (A'B') \cap (XP)\).
Both angles \(XAA'\) and \(XBB'\) are right; therefore
\(2 \cdot \measuredangle XAA' \equiv 2 \cdot \measuredangle XBB'\).
By Corollary 9.3.2 , \(\square XAPB\) is inscribed. Applying this theorem again we get that
\(2 \cdot \measuredangle AXP \equiv 2 \cdot \measuredangle ABP\).
Since \(\square ABA'B'\) is inscribed,
\(2 \cdot \measuredangle ABB' \equiv 2 \cdot \measuredangle AA'B'\).
It follows that
\(2 \cdot \measuredangle AXY \equiv 2 \cdot \measuredangle AA'Y\).
By the same theorem \(\square XAYA'\) is inscribed, and therefore,
\(2 \cdot \measuredangle XAA' \equiv 2 \cdot \measuredangle XYA'\).
Since \(\angle XAA'\) is right, so is \(\angle XYA'\). That is \((XP) \perp (A'B')\).
Find an inaccuracy in the solution of Problem \(\PageIndex{1}\) and try to fix it.
- Hint
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One needs to show that the lines \((A'B')\) and \((XP)\) are not parallel, otherwise the first line in the proof does not make sense.
In addition, we implicitly used the following identities:
\(2 \cdot \measuredangle AXP \equiv 2 \cdot \measuredangle AXY\), \(2 \cdot \measuredangle ABP \equiv 2 \cdot \measuredangle ABB'\), \(2 \cdot \measuredangle AA'B' \equiv 2 \cdot \measuredangle AA'Y\).
The method used in the solution is called method of additional circle, since the circumcircles of the quadrangles \(XAPB\) and \(XAPB\) above can be considered as additional constructions.
Assume three lines \(\ell, m\), and \(n\) intersect at point \(O\) and form six equal angles at \(O\). Let \(X\) be a point distinct from \(O\). Let \(L, M\), and \(N\) denote the foot points of perpendiculars from \(X\) on \(\ell, m\), and \(n\) respectively. Show that \(\triangle LMN\) is equilateral.
- Hint
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By Corollary 9.3.1 , the points \(L, M\), and \(N\) lie on the circle \(\Gamma\) with diameter \([OX]\). It remains to apply Theorem 9.2.1 for the circle \(\Gamma\) and two inscribed angles with vertex at \(O\).
Assume that a point \(P\) lies on the circumcircle of the triangle \(ABC\). Show that three foot points of \(P\) on the lines \((AB), (BC)\), and \((CA)\) lie on one line. (This line is called the Simson line of \(P\)).
- Hint
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Let \(X, Y\), and \(Z\) denote the foot points of \(P\) on \((BC), (CA)\), and \((AB)\) respectively. Show that \(\square AZPY\), \(\square BXPZ\), \(\square CYPX\), and \(\square ABCP\) are inscribed. Use it to show that
\(\begin{array} {rcl} {2 \cdot \measuredangle CXY \equiv 2 \cdot \measuredangle CPY} & \ \ \ \ & {2 \cdot \measuredangle BXZ \equiv 2 \cdot \measuredangle BPZ,} \\ {2 \cdot \measuredangle YAZ \equiv 2 \cdot \measuredangle YPZ} & \ \ \ \ & {2 \cdot \measuredangle CAB \equiv 2 \cdot \measuredangle CPB.} \end{array}\)
Conclude that \(2 \cdot \measuredangle CXY \equiv 2 \cdot \measuredangle BXZ\) and hence \(X, Y\), and \(Z\) lie on one line.