Question

In given fig. AB and CD are common tangents to two circles of unequal radii. Prove that AB=CD.

Answer 1

construction:produce the tangents upto point p
now let p be the point of intersection
the length of two tangents drawn frm any external point to a circle r equal
ap=cp_1st equation
bp=dp_2nd equation
add 1 & 2
we get:ap+bp=cp+dp
p nd p are cancelled
so ab=cd
h.p

Answer 2

EXTEND AB TO P AND CD TO P AS SHOWN IN FIGURE

WKT TWO TANGENTS DRAWN FROM AN EXTERNAL POINT ARE EQUAL SO PB=PD

AND AP=PC

SUBTRACTING THE SMALLER TANGENTS FROM BIG TANGENTS

AP-PB=PC-PD

AB=CD

HENCE PROVED