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This solution comes from Sue Liu of Madras College
Let the large semicirle have diameter $AB$ and centre $X$.
Let the two smaller semicircles have centres $C$ and $D$ and radii $R$ and $r$.
Thus $AC = R$, $BD = r$ so that $AX = (2R + 2r)/2 = (R + r)$ and $CX = r$ from which it follows that $XD = R$.
Let the small circle have have centre $O$ and radius $x$.
Then $CO = R + x$ and $DO = r + x$.
The line $XO$, joining the centre of the large semicircle to the centre of the small circle, cuts the circumference of the large semicircle at $E$ where $XE = XB = R + r$, $OE = x$ and $OX = R + r  x$.
If we now consider the triangle $OCD$ we have

$\angle OXC + \angle OXD = 180^o$
So
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