In right triangle ΔABC (m∠C = 90°), point P is the intersection of the angle bisectors of the acute angles. The distance from P to the hypotenuse is equal to 4 in. Find the perimeter of △ABC if AB = 20 in.

The intersection point of the angle bisectors is the incenter of the triangle, the center of the incircle, which has radius 4.

One of the properties of the incircle is that the distances (d) from vertex C to the nearest touchpoints are equal and have the value
  d = 1/2(a +b -c)
where a, b, c are the lengths of the sides opposite angles A, B, C, respectively.

We know the distance from the right-angle vertex C to the incircle touchpoints is the radius of that circle, 4 in. And, we know the length of the hypotenuse (c) is 20 in. This lets us find the sum a+b as ...
  4 = (1/2)(a+b -20) . . .. fill known information into the formula
  8 = (a+b -20) . . . . . . . multiply by 2
  28 = a+b . . . . . . . . . .. the total length of the two legs of the triangle

The perimeter is the sum of the leg lengths and the hypotenuse length.
  perimeter = 28 in + 20 in = 48 in

The perimeter of ΔABC is 48 in.


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It turns out that the triangle is a 3-4-5 triangle scaled by a factor of 4, so having sides 12, 16, and 20.