Question

in a right angled triangle ABC ,angle ABC is 90°,A circle is inscribed in the triangle with radius r,a,b,c are the length of side BC,AC,and side AB respectively then,Prove that:2r=a+b-c

Answer 1

Step-by-step explanation:

since these points produce tangents at the circle, hence

AP = AQ

CP = CR

BR = BQ

Given,

AB = c

AC = b

BC = a

since CPOR is a square hence,

CP = CR = r

=> BR = BC - CR = a - r

=> BQ = BR = a - r

similarly,

AP = AC - CP = b - r

=> AQ = AP =  b - r

Now,

AB = c

=> AQ + BQ = c

=> b - r + a - r = c

=> b + a - 2r = c

=> 2r = a + b - c

which is the required equation.

Answer 2

Answer:

since these points produce tangents at the circle, hence

AP = AQ

CP = CR

BR = BQ

Given,

AB = c

AC = b

BC = a

since CPOR is a square hence,

CP = CR = r

=> BR = BC - CR = a - r

=> BQ = BR = a - r

similarly,

AP = AC - CP = b - r

=> AQ = AP =  b - r

Now,

AB = c

=> AQ + BQ = c

=> b - r + a - r = c

=> b + a - 2r = c

=> 2r = a + b - c

which is the required equation.

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