Question

Let S denote the semi-perimeter of a triangle ABC in which BC = a, CA = b, AB = c.If a circle touches the sides BC, CA, AB at D, E, F respectively, Prove that BD = s -b.

Answer 1

Step-by-step explanation:

Solution :

A circle is inscribed in the ΔABC, which touches the BC, CA and AB.

Given, BC=a, CA=b and AB=c

By using the property, tangents are drawn from an external point to the circle are equal in length.

∴BD=BF=x [say]

DC=CE=y [say] and AE=AF=z [say]

Now, BC+CA+AB=a+b+c ⇒(BO+DC)+(CE+EA)+(AF+FB)=a+b+c

⇒(x+y)+(y+z)+(z+x)=a+b+c

⇒2(x+y+z)=2s

[∴2s=a+b+c=perimeter of ΔABC]

⇒s=x+y+z

⇒x=s−(y+z)

⇒BD=s−b [∵b=AE+EC=z+y]

Hence proved.

Answer 2

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