AB is a line segment and M is its mid point.Semicircles are drawn with AM,MB and AB as diameters on

Asked by aditi.priya | 5th Oct, 2009, 03:04: PM

Expert Answer:

 

 

 

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.

 

Answered by  | 7th Oct, 2009, 04:13: PM

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