If C is the mid point of the line segment AB . L and M are the mid point of the line segment AC and BC respectively
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Let AB = t units.
Let L, N be the mid points of AM,MB respectively.
Let C (0, ) be the given circle which touches the semi circles with centres L,
M, N at P, R and Q respectively.
Now join OL and ON.
The points O, P, L are collinear and the points O, Q, N are collinear.
By similarity points M, O, R are collinear
Here AL = LM = MN = NB = t/4 units.
[since M is the mid point of AB, L is the mid point of AM and N is the mid point of MB]
Also OL = OP + LP
= r + t/4 [since LP = t/4]
And ON = OQ + QN
ON = r + t/4
Þ ΔOLN is an isosceles triangle and M is the mid point of the base LN.
Þ OM ^ LN
\ OMN is a right angled triangle.
Thanks
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