a circle is inscribed in a triangle ABC having sides AB=10 cm AC=8cm BC=12cm such that it touches AB at F AC at E and BC at D.
find AF , BD and CE
Tangents drawn from an external point to a circle are equal.
Let
AD = AF = x
BD = BE = y
and CE = CF = z
Now,
AB = AD + DB = x + y = 10 ...(1)
BC = BE + EC = y + z = 12 ...(2)
AC = AF + FC = x + y = 8 ...(3)
Adding (1), (2) and (3), we get,
⇒2 (x + y + z) = 30
⇒ (x + y + z) = 15 ...(4)
On Solving (1) and (4), we get, z =5
Solving (2) and (4), we get x = 3
Solving (3) and (4), we get y = 7
Hence,
AD = AF= x = 3 cm
BD = BE = y = 7 cm
CE =CF = z = 5 cm