MN is the common chord of two intersecting circles and AB is their common tangent. prove that the line NM produced bisects AB at point P.
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According to question,
AB is the common tangent.
MN is the common chord.
common chord MN touch with common tangent AB at P.
we know,
square of tangent of circle is equal to product of length of secant and length of its exterior secant . Use this therem,
so,
AP² = PN.PM ----(1)
BP² = PN.PM-----(2)
from equations (1) and (2),
AP² = BP² => AP = BP
It is now clear that P is the midpoint of AB.
Hence, P bisects AB.
AB is the common tangent.
MN is the common chord.
common chord MN touch with common tangent AB at P.
we know,
square of tangent of circle is equal to product of length of secant and length of its exterior secant . Use this therem,
so,
AP² = PN.PM ----(1)
BP² = PN.PM-----(2)
from equations (1) and (2),
AP² = BP² => AP = BP
It is now clear that P is the midpoint of AB.
Hence, P bisects AB.
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