Semicircular paper and creasing of a chord A semicircular piece of paper with radius $2$ $cm$ is folded along a chord so that the arc is tangent to the diameter.If the contact point of the arc divides the diameter in the ratio $3:1$,determine the length of the crease.
Sorry I can't show any work as I can't develop any approach to solve this question. Any suggestions?
Picture:

 A: Hint: Let the  semicircle be centered at the origin with radius 2, and lie above the x axis, so that the point $P=(1,0)$ cuts the diameter in a 3:1 ratio. Then consider another circle also of radius 2 but centered at $(1,2)$ so it would touch the x axis (diameter) at P so be tangent to the diameter there. 
Figure out by algebra the two points where the new circle goes through the original semicircle, and get the length of the chord via distance formula.
A: 
In the figure, arc AB touches MN at P. K, the center of that arc, lies somewhere on PH where $PH \bot MN$.
Arc AB is a part of the image of the circle AMNB. K is therefore located at a distance of 2 units from P.
In the brown triangle, $OK = \sqrt 5$.
From the $[\theta, 2, 2, \sqrt 5]$ green triangle, $\theta$ can be found by cosine law.
From the $[2\theta, 2, AB, 2]$ triangle KAB, AB can be found by cosine law again.
A: Let $O$ be the center of the semicircle (the red dot),
$P$ the point of tangency of the folded segment, and let $A$ and $B$
be the ends of the "crease" as shown in the figure below:

The fold makes a mirror image of the arc from $A$ to $B$ across the line $AB$.
Everything about that arc is mirror-imaged, including the location of its
center of curvature. That is, the triangle $\triangle OAB$
is reflected to $\triangle QAB$, where $Q$ is the center of curvature of
the reflected arc.
(You can think of the entire "pie slice" $OAB$, bounded by the red lines
and the original semicircle, as being flipped over without moving the line
segment $AB$, to produce the "pie slice" $QAB$.)
The clue to the size of $\triangle OAB$ and $\triangle QAB$
is the point of tangency. Since $Q$ is the center of curvature of the folded
arc, the radius $QP$ has to be perpendicular to the original
semicircle's diameter at $P$ (the point of tangency),
and since the original arc had radius $2$, so does its mirror image,
so $QP = 2$.
Since $P$ divides the diameter (of length $4$) in the ratio $3:1$,
it is $1$ unit from the end of the diameter, and therefore $OP = 2 - 1 = 1$.
So now we have two legs of the right triangle $\triangle OPQ$,
and we can find the length of the hypotenuse $OQ$.
Knowing $OQ$ and using the symmetry of the reflection, we can
determine the length $QR$. Knowing that $QA = OA = 2$ and $QB = OB = 2$
(because of the radius of the original circle, and the reflection due
to folding), we now have the hypotenuse and one leg of $\triangle ARQ$
and can find the length of the remaining leg.
The result for $\triangle BRQ$ is the same.
And those two results together tell us the length of $AB$.
A: 
Calculation has two parts. Finding the segment on outside circle and the angle  $ \alpha $ subtended between fold/diameter.
By creasing or folding T falls on t
$$ r = nK = 2 ( 2 cm) ;\, OK = k ;$$
$$(k+r) ( k-r ) = OT^2 = Ot^2 = ( k-r/2)^2 \rightarrow  k = 5 r /4 $$
$$ on = om = OK- nK= r/4 ; \,  OT = Ot= on + nt= r/2 + r/4 = 3r/4 $$
dot marked angle $= \alpha , $
$$  \tan 2 \alpha = \ tan TOK  = TK/OT = 1/(3/4) = 4/3 $$
$$ \cos 2 \alpha = 3/5 ;\, \cos \alpha = 2/\sqrt 5 ;\, $$
$$ OM = 5 r/4 \cdot 2/\sqrt 5 = \sqrt 5 r/2 ; $$
By Circle segment product property,
$$ OT^2 = OP \cdot OQ = (OM + MQ) (OM-MP) = OM^2 - x^2 $$ 
$$ (if \, MP = MQ = x) $$
$$ x^2 = OM^2 - OT^2  = 5 r^2/4 - (3 r/4)^2 ;\, \rightarrow x = \sqrt {11}/4 $$
$$ PQ = 2 x  = r \sqrt{11}/2 = \sqrt{11} cm \;  (Answer).\,$$
