circleseuclidean-geometry
I found this mathematical expression for the length of a arc $l(r)$ i.e the shorter arc ACB. In other words, why is $l(r)$ equal to that expression with respect to R i.e $l (r) = 2r \arccos (r/2R)$? I have tried hard to prove it but I couldn't. I hope someone could give the clue to the answer. The diagram and question can be found the the attached picture
Let $X$ be the point where lines $BC$ and $FG$ meet. Let $J$ be the point where BC intersects $AE$. Let's forget about point $I$ for a moment. Here is the diagram:
Observe that $BD$ is the bisector of angle $HBA$, therefore $HB/BA=HD/DA$. Similarly, $HC/CA=HD/DA$. We see that points $B$, $C$ and $D$ all have the same ratio of distances from points $H$ and $A$. It follows that $(O)$ is an Apollonian circle of segment $HA$. Since $E$ is on this circle too, $HE/EA=HD/DA$. It follows that $ED$ is the bisector of angle $HEA$.
Now let us have a look at lines $EX$, $EB$, $EJ$, $EH$ and $EC$ and some of their cross-ratios. Let $f$ be the reflection across line $EX$. What we have established above is that $f(EJ)=EH$. It is also an easy exercise to see that $f(EB)=EC$ and $f(EX)=EX$. So, $f$ maps the four lines $EJ$, $EB$, $EC$ and $EX$ to lines $EH$, $EC$, $EB$ and $EX$ respectively. It follows that the corresponding cross-ratios are the same: $$ (EJ,EB;EC,EX) = (EH,EC;EB,EX). $$ Therefore, we have an equality for cross-ratios of points on line $BC$: $$ (J,B;C,X) = (H,C;B,X). $$
If we look at the perspective projection from line $BC$ to line $FG$ with center $A$, it sends points $J$, $B$, $C$ and $X$ to $E$, $F$, $G$ and $X$ respectively. Since perspective projections preserve cross-ratios, we have: $$ (E,F;G,X) = (H, C; B, X). $$
Now let us look at point $I$ (not shown on my diagram) where $BG$ and $FC$ intersect. Let $g$ be the perspective projection from line $BC$ to line $FG$ with center $I$. It is clear that $g(X)=X$, $g(B)=G$ and $g(C)=F$. $g$ preserves cross-ratios, therefore $$ (H, C; B, X) = (g(H), F; G, X), $$ and so $$ (E,F;G,X) = (g(H), F; G, X). $$ It follows that $g(H)=E$, which means that $H$, $E$ and $I$ are collinear, QED.
One way to define the arc length of a curve $\gamma:[a,b]\to\mathbb R^2$ is by considering partitions of $[a,b]$, as in Riemann integration (read this Wikipedia article first): if $P(x,t)$ is a tagged partition of $[a,b]$, then $L_\gamma(P)$, the length of $\gamma$ with respect to the partition $P$, is $\sum_{i=1}^{n-1}|\gamma(t_i)-\gamma(t_{i-1})|$.
Definition: If $\underset{P}{\lim \sup} L_\gamma(P)$ exists, then $L_\gamma = \underset{P}{\lim \sup} L_\gamma(P)$ is the length of the curve.
Now consider the portion of the red arc between the point $A$ and the line $BD$. Using the above definition, we can see that the length of this arc lies between $AD$ and $AB$: if $\gamma(t_{i-1})$ and $\gamma(t_i)$ are points on the curve, we can project them perpendicular to $AD$ onto the points $r_{i-1}$ and $r_i$ on $AD$, and $s_{i-1}$ and $s_i$ on $AB$. And then we have $|r_i-r_{i-1}| < |\gamma(t_i)-\gamma(t_{i-1})| < |s_i-s_{i-1}|$, by simple geometry.
Therefore $L_{AD}(P) < L_\gamma(P) < L_{AB}(P)$ (taking some liberties with the notation). And in the limit we get:
$$\underset{P}{\lim \sup}L_{AD} \le \underset{P}{\lim \sup}L_\gamma(P) \le \underset{P}{\lim \sup}L_{AB}$$
In other words: $$AD \le L_\gamma\le AB$$
To show strict inequality would not be too difficult $-$ for instance, if we are more than halfway to the line $BD$, then there is a constant $\mu < 1$ such that $|\gamma(t_i)-\gamma(t_{i-1})| < \mu|s_i-s_{i-1}|$. This would give $L_\gamma(P) < \frac12(1+\mu)L_{AB}(P)$, so $L_\gamma \le \frac12(1+\mu)AB$.
Best Answer
We have
$$ l(r) = 2r \angle ADC. $$
So we just need to compute $\angle ADC$. Let $Q$ be the center of the circle on the left. Note that $AD=r$ and $DQ=AQ=R$. It follows that $\angle ADQ=\angle ADC=\arccos(r/2R)$. To see why, note that $\triangle ADQ$ is isosceles and drop a perpendicular from $Q$ to the midpoint of $AD$.
Is that clear?