A circle is inscribed in triangle ABC that touches the sides AB, BC, CA at points P, Q, R

A circle is inscribed in triangle ABC that touches the sides AB, BC, CA at points P, Q, R find AP, PB, BQ, QC, CR, RA, if AB = 10 cm, BC = 12 cm, CA = 5 cm.

Since the lengths of the segments of tangents drawn to the circle from one point are equal, then AR = AP, BP = BQ, CQ = CR.

For convenience, we denote pairwise equal segments AR = AP = X, BP = BQ = Y, CQ = CR = Z.

Then:

AB = X + Y = 10. (1).

AC = X + Z = 5. (2).

BC = Y + Z = 12. (3).

Let’s solve the system of their three equations by the addition method.

Let us subtract the second from the first equation.

(X + Y) – (X + Z) = 10 – 5.

Y – Z = 5.

Let’s add the third equation to the last one.

(Y + Z) + (Y – Z) = 12 + 5.

2 * Y = 17.

Y = 17/2 = 8.5 cm.

Plug in the Y value and find X and Z.

X + 8.5 = 10.

X = 10 – 8.5 = 1.5 cm.

Z = 12 – Y = 12 – 8.5 = 3.5 cm.

Then: AR = AP = 1.5 cm, BP = BQ = 8.5 cm, CQ = CR = 3.5 cm.

Answer: AR = AP = 1.5 cm, BP = BQ = 8.5 cm, CQ = CR = 3.5 cm.