Mon November 21, 2011 By:

C1 and C2 are two concentric circles,the radius of C2 being twice that of C1.From a pomt P on C2,tangents PA and PB are drawn to C1.Prove that the centroid of the triangle PAB lies on C1.

Expert Reply
Wed November 23, 2011
See in the above figure,
OP is the radius hence OP=2r
In triangle OAP, since PA is tangent and OA is radius it means
angle OAP=90
So, OA=r OP=2r, now using pythagoras theorem in triOAP
we get AP=?3r
Similarly PB=?3r
So triPAB is isosceles
Now in tri OAP
angle APO= tan-1(r/?3r) = 30 
similarly angle BPO=30
hence angle APB=30+30=60
now tri APB is isosceles means angle PAB=angle PBA = 60
hence triangle PAB is equilateral triangle.
Now QP= OP-OQ = 2r-r = r
In tri OAQ angle QOA=60 and OA=OQ=r, means this triangle is also equilateral
so AQ=r
similarly QB=r
hence we have QA=QB=QP=r means Q is equidistant from the vertices
of the equilateral triangle PAB, hence Q is centroid. since for an equilateral
triangle, centroid is equidistant from all the vertices.
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