Some circles and a constant segment.

freemind

#1

Some circles and a constant segment.

**Posted:** Sun Mar 30, 2008 3:45 pm

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Clark Kent · New Member Offline Joined: 13 Feb 2008 Posts: 12

#2

Re: Some circles and a constant segment.

**Posted:** Sun Mar 30, 2008 5:38 pm

pohoatza

#3

Denote by $X$ the intersection point of the perpendiculars in $M$ , $N$ to the lines $DM$ , and $DN$ , respectively. Since the quadrilaterals $DMXN$ and $MCND$ are cyclic, we have that $\angle{MXD} = \angle{MCD} = \angle{C}/2 = \angle{DCN} = \angle{DXN}$ . The triangles $DMX$ and $DNX$ are now congruent, and therefore the segments $XM$ , and $XD$ are equal. Thus, the circle with center $X$ and radius $XM = XD$ is tangent to the lines $DM$ , $DN$ at points $M$ , $N$ , respectively. This proves a).

Next, notice that the triangles $MXC$ and $QXC$ are congruent (this is because $XM = XQ$ , $\angle{XMC} = \angle{XNC} = \angle{XQC}$ , and $XQ$ is common), and thus $MC = QC$ . Similarly, the triangles $XCP$ and $XCN$ are congruent, and hereby $CP = CN$ . Since $MP = NQ$ , it is suffice to prove that $MC + CN$ is constant. But this follows from Ptolemy's theorem, applied to the quadrilateral $MCND$ , since
$MC + CN = CD \cdot \frac {MN}{MD}$
(we know that $MD = DN$ from the congruence of the triangles $DMX$ and $DNX$ ), and $\frac {MN}{MD} = 2\cos{\frac {C}{2}}$ . This proves b).

**Posted:** Sun Mar 30, 2008 5:53 pm

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