A circle is inscribed in a triangle ABC such that it touches the sides BC, CA and AB at D, E and F respectively as shown in the figure. If BC = 5 cm and AB + AC = 13 cm, then AE equals
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We know that tangent drawn from external point to a circle are equal
So AD=AF=x
Or BD=BE=y
And CF=CE=z
Now, AB=AD+DB=x+y=12 cm ...(1)
Or BC=BE+EC=y+z=8 cm ...(2)
And AC=AF+CF=x+z=10 cm ...(3)
Adding the above three equation, we get
2(x+y+z)=12+8+10=30 cm
Or x+y+z=15 cm
As per equation (1),
x+y=12 cm
Then, z=15−10=5 cm =CF
As per equation (3),
x+z=12 cm
Then, y=15−12=3 cm =BE
As per equation (2),
y+z=8 cm
Then, x=15−8=7 cm =AD
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