If parallel chords of a circle subtend central angles of $72^\circ$ and $144^\circ$, then the perpendicular distance between them is half the radius I'm not much fond of proving results when it comes to geometry; still, being that I highly admire the subject, I believe I am getting better at it as time so goes by. Yet, one some problem has seemed so such that it got the best of me. The problem is dictated as so:

Two parallel chords of a circle, subtend angles of $72^\circ$ and $144^\circ$ respectively at the centre. Prove that the perpendicular distance between the chords is half the radius of the circle.

The way I initially thought of approaching this problem is by dropping this so required perpendicular line from the point of interection of the first chord with the circumference of the circle upon the second chord (being that the first chord is implicitly above the second). The length of the perpendicular is to be equated to some relation it holds with the radius of the circle (namely, this relation is that it is half of the radius). Any rule I tried to apply, or any law I thought of executing ultimately led me to a dead end.
I do believe that the solution of this problem is to be obtained using one of the trigonometric sum/difference to product rules, since the problem appears in a trigonometry textbook under that said chapter.
Any amount of help would be very appreciated. It would be appreciated even more if the reader could also share their train of thought and reasoning when they usually tackle problems in the likes of this (since I am still horribly new in making geometric proofs other than those shown in Euclid's Elements).
Moreover, it would also be quite the helpful if you could educate me on how such proofs are supposed to be carried out in general, and the general train of thought that should be followed to get results more efficiently. This problem is present in S.L. Loney's book on plane trigonometry, a wondeful book for getting the big picture of trigonometry, if anyone would be of the interest. Thank you in advance.
 A: A fast track using congruences
Let $AB$ and $CD$ be the two parallel chords in the hypothesis, and let $O$ be the circle center. Complete the graph as indicated below. In particular let $E$ be the midpoint of the smaller of the two arcs $AB$, $F$ the intersection between $CD$ and $OE$, $G$ the intersection between $AD$ and $BC$, and $H$ the intersection between $AB$ and $OE$.

Note that $A$, $B$, $C$, $D$, and $E$ are verteces of the regular decagon inscribed in the circumference. Use this fact, and angle chasing, to prove that $OCGD$ and $AEBG$ are rhombuses, and that $G$ lies on $OE$. Then use the properties of the rhombus's diagonals to show that $OF \cong FG$ and $GH\cong HE$, hence the thesis.

As for your request on a possible way of thinking, that really is subjective. In this case I personally wanted to try avoiding trigonometry. Nothing bad in using it, but as a high school teacher I must often reduce the available instruments to the mimimum, which is, I believe, also a good way to dig deeper into the subject.
That said, in this case I noticed the repeated $36^\circ$ angle, but I was at the same time particularly annoyed by the fact that the segment $FH$ (which we wanted to prove being half the radius) did not share any end point with an actual radius. Thus I tried to add as few lines as possible that could translate this "halving" situation into a clearer scenario. And that is how I ended up with the above construction. I leave all the details to you.
A: 
Let $O$ the center of the circle, $r$ your radius and $AB$ e $CD$ two parallel chords. Let $A\hat O B=72°$ and $A\hat O M=36°$. We know that $ON\perp CD$ and $OM\perp AB$. In the triangle $\triangle A O M$ ($A \hat M O=90°$) the angle $$ O \hat A M=180°-A \hat M O-A \hat O M=54°$$
In the triangle $\triangle A O M$
$$\sin 54°=\frac{|OM|}{r}\implies |OM|=r\sin 54°$$
The angle $C \hat O D=144°$ then $N \hat O D=72°$ and $O \hat D N=180°-90°-72°=18°$. $|OD|=r$ and thus $|ON|=r\sin 18°$.
But
$$|OM|-|ON|=r\sin 54°-r\sin 18°=r\left[\left(\frac{\sqrt5+1}4\right)-\left(\frac{\sqrt5-1}4\right)\right]=\frac r2$$
A: The distance from a chord intercepting an arc $\theta<180°$ equals $r\cos(\theta/2)$, so you really are aiming to prove that
$\cos(36°)-\cos(72°)=(1/2).$
If we know that the cosines have values $(\pm1+\sqrt5)/4$, then we can just plug and chug. Here a method is presented that does not require this prior knowledge.
Let $x=\cos(36°)-\cos(72°)$ and seek $x^2$:
$x^2=\cos^2(36°)-2\cos(36°)\cos(72°)+\cos^2(72°).$
We apply the product-sum relation $\cos(u)\cos(v)=(1/2)[\cos(u+v)+\cos(u-v)]$:
$x^2=(1/2)[\cos(72°)+1]-[\cos(108°)+\cos(36°)]+(1/2)[\cos(144°)+1].$
Apply the supplementary-angle relation $\cos(180°-u)=-\cos(u)$ for the terms with $108°$ and $144°$. You then have only the cosines of $36°$ and $72°$ remaining and can combine like terms. This ultimately gives:
$x^2=(-3/2)[\cos(36°)-\cos(72°)]+1.$
The expression in brackets is just $x$, so we have
$x^2=(-3/2)x+1,$
which can be solved as a typical quadratic equation. One root is $x=-2$, which has both the wrong sign ($x$ should he positive) and the wrong magnitude for the geometric construction (two radii equals the whole diameter of the circle, not the distance between two parallel chords). Thus the other root $x=1/2$ is forced to hold, qed.
The big reveal
During the above analysis we render
$\cos(36°)\cos(72°)=(1/2)[\cos(108°)+\cos(36°)]$
from the trigonometric sub-product relation. But also, $\cos(108°)=-\cos(72°)$, so the bracketed expression above is just $\cos(36°)-\cos(72°)=(1/2)$. So we now have
$\cos(36°)-\cos(72°)=(1/2)$
$\cos(36°)\cos(72°)=(1/4)$
So $\cos(36°)$ and $-\cos(72°)$ have sum $1/2$ and product $-1/4$, whereupon they must be the roots of the quadratic equation
$4y^2-2y-1=0.$
By solving this equation with the quadratic formula, the precise values of the two cosines which were supposed to be unknown are in fact revealed!
A: If you drop a perpendicular from the center onto the longer (shorter) chord, its length will be $r\cos72^{\circ}$ $(r\cos36^{\circ})$ respectively and thus the distance between them is $(\cos36^{\circ}-\cos72^{\circ})r$, since the shorther chord is farther. The answer is $\frac{1}{2}r$.
Proof: It is enough to know that $\cos36^{\circ}=\frac{\phi}{2}$ where $\phi$ is the Golden ratio, since then I can calculate $\cos 72^{\circ}=2\cos^236^{\circ}-1=\frac{\phi^2-2}{2}=\frac{\phi+1-2}{2}=\frac{\phi-1}{2}$ and $\cos36^{\circ}-\cos72^{\circ}=\frac{\phi}{2}-\frac{\phi-1}{2}=\frac{1}{2}.$
A: Consider symmetry. If tangent to circle at two segment ends rotates $\phi$ w.r.t. horizontal anti clockwise direction and
$\theta$ the central angle, then for a  circle radius $a$
$$ \phi = \theta+ \pi/2$$
$$ y= a \sin \theta = -a \cos \phi $$
Evaluate the difference of $y$ coordinate directly at two slope points by cosine difference trigonometric formula in the given  particular case:
$$ (y_2- y_1) =a(\cos 36^{\circ}-\cos 72^{\circ})= ~2 a\sin 54~~\sin 18{^\circ}=~2a \frac{\sqrt 5 +1}{4}\frac{\sqrt 5 -1}{4} = \frac{a}{2}. $$
A: 
In the figure, $EF$ and $DG$ are perpendicular to $AC$. Our job is to prove that $FH=\frac{OC}{2}$.
Note that $G$ is the point where $OF=FG$. $ED$ and $OC$ are extended to meet at $I$ as shown.
We are done if we can prove that $GH=HC$.
(1) $\Delta ODE \cong \Delta EOG$ (ASA)
(2) Hence $OG=DE$
(3) $\Delta IOE$ is an isosceles triangle and $OG=DE \implies IG=ID$
(4) Hence $\angle HGD= \frac{180^o-36^o}{2}=72^o$
(5) $\Delta DHG \cong \Delta DHC$  (AAS)
(6) $\therefore GH=HC$
Job done.
