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# 9.4: Method of additional circle

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Problem $$\PageIndex{1}$$

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 $$(XP) \perp (A'B')$$.

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 $$(XP) \perp (A'B')$$.

Exercise $$\PageIndex{1}$$

Find an inaccuracy in the solution of Problem $$\PageIndex{1}$$ and try to fix it.

Hint

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.

Exercise $$\PageIndex{2}$$

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

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 $$O$$.

Advanced Exercise $$\PageIndex{3}$$

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

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.