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$2$. Let $ABC$ be an acute-angled triangle, and let $D$,$E$ be the feet of the perpendiculars from $A$, $B$ to $BC$, $CA$ respectively. Let $P$ be the point where the line $AD$ meets the semicircle constructed outwardly on $BC$, and $Q$ be the point where the line $BE$ meets the semicircle constructed outwardly on $AC$. Prove that $CP = CQ$.

I've haven't made much progress at all. Can anyone give a complete proof. Thanks in advance.

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Relative to an origin O let the position of A be $\underline a$, and so on.

From the given information we have $$(\underline p-\underline a).(\underline b-\underline c)=0$$ $$(\underline b-\underline q).(\underline c-\underline a)=0$$

The angle in a semicircle is a right angle also gives us $$(\underline p-\underline c).(\underline b-\underline p)=0$$ $$(\underline q-\underline c).(\underline a-\underline q)=0$$

We need to show that $$(\underline c-\underline p).(\underline c-\underline p)=(\underline c-\underline q).(\underline c-\underline q)$$ which is equivalent to showing $$\underline p.\underline p-\underline q.\underline q=2(\underline c.\underline p-\underline c.\underline q)$$

To get this result, expand the third and fourth equations and subtract them. This gives $$\underline p.\underline p-\underline q.\underline q=\underline c.\underline p-\underline c.\underline q+[\underline p.\underline b-\underline c.\underline b+\underline a.\underline c-\underline a.\underline q]$$

However, if we expand the first two equations and subtract them we also get, after rearrangement, $$\underline p.\underline b-\underline c.\underline b+\underline a.\underline c-\underline a.\underline q=\underline c.\underline p-\underline c.\underline q$$ So the required result follows immediately

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  • $\begingroup$ What are you actually referring to by position, the $x$ co-ordinate, or the distance, or something else. $\endgroup$ – MadChickenMan Jun 23 '15 at 12:02
  • $\begingroup$ $\underline a$ is the vector $\vec {OA}$ and so on $\endgroup$ – David Quinn Jun 23 '15 at 12:11
  • $\begingroup$ I think I get it all now, thanks. I was trying to use basic angle/circle theorems, but I'm not sure there is an easy proof using that method, though I may just have been missing something. $\endgroup$ – MadChickenMan Jun 23 '15 at 12:44
  • $\begingroup$ There probably is a geometry proof, but I couldn't see it. The vector approach seems fairly straightforward. I wonder if there's a nice way of doing it with complex numbers? $\endgroup$ – David Quinn Jun 23 '15 at 12:46
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enter image description here

$\triangle BPC$ is a right triangle, and $\triangle PDC \sim \triangle BPC$, so

$$\frac{|\overline{CD}|}{|\overline{PC}|} = \frac{|\overline{PC}|}{|\overline{BC}|}\qquad \to\qquad |\overline{PC}|^2 = |\overline{BC}|\;|\overline{CD}|=|\overline{BC}|\;|\overline{AC}|\;\cos C \quad (\star)$$

Since the right-hand side of $(\star)$ is symmetric in $A$ and $B$, we'd get the same result for the point $Q$ constructed via the perpendicular from $B$ to $\overline{AC}$. Therefore, $|\overline{PC}| = |\overline{QC}|$. $\square$


Note: There's nothing special about $P$ and $Q$ being on the "outer" semi-circles. The circle about $C$ through $P$ (or $Q$) passes through all four points where the altitudes from $A$ and $B$ meet the circles having the opposite edges as diameters.

enter image description here

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  • $\begingroup$ Woops, I was actually trying to prove the case of an inwardly facing semicircle. Thanks for your answer anyway, nice and simple. $\endgroup$ – MadChickenMan Jun 23 '15 at 13:13

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