Answer:
CR = 17 ; PR = 30
Explanation:
From the problem hypothesis we know that CD⊥PR or CQ⊥PR(perpendicular) and ∡PCQ ≡∡RCQ (bisector)
so ∡PCQ≡∡RCQ
[CQ]≡[CQ] (common side)
CQ⊥PR
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⇒(cathetus - angle case of congruence)⇒ ΔPQC≡ΔRQC so [PQ]≡[QR] (=15) // PR = PQ+QR ⇒ PR=30
from this congruence ⇒ΔPRC = isosceles so PC=CR=17
Hint: in any triangle is bisector of an angle it is perpendicular on the opposite side then triangle is isosceles.