Samy1234
Let AD = x thus, BD = 12-x now, BE = BD = 12-x (tangents from an external point) or, CE = 8-(12-x) = 8-12+x = x-4 again, CE = CF = x-4 (tangents from an external point) or, AF = 10-(x-4) = 10-x+4 = 14-x But, AF = AD = x (tangents from an external point)
BTP, x = 14-x or, 2x = 14 or, x = 7 or, AD = 7cm BE = 5cm CF = 3cm
Answers & Comments
thus, BD = 12-x
now, BE = BD = 12-x (tangents from an external point)
or, CE = 8-(12-x)
= 8-12+x = x-4
again, CE = CF = x-4 (tangents from an external point)
or, AF = 10-(x-4) = 10-x+4 = 14-x
But, AF = AD = x (tangents from an external point)
BTP,
x = 14-x
or, 2x = 14
or, x = 7
or, AD = 7cm
BE = 5cm
CF = 3cm