Answer: B
In triangle ABC, the incenter is I. ∴ AI and BI are the internal bisectors of ∠BAC and ∠ABC respectively. In triangle ABI, the exterior angle ∠BIP = ∠IAB + ∠IBA = \(\frac{1}{2}\angle\)BAC + \(\frac{1}{2}\angle\)ABC = \(\frac{1}{2}[\angle\)BAC + ∠ABC] Again, ∠IBP = ∠IBC + ∠CBP = \(\frac{1}{2}\angle\)ABC + ∠PAC (∵ both are angles subtended by arc PC on the circle) = \(\frac{1}{2}\angle\)ABC + \(\frac{1}{2}\angle\)BAC ∴ In triangle IBP, ∠BIP = ∠IBP ∴ PI = PB = 15 cm
In triangle ABC, the incenter is I. ∴ AI and BI are the internal bisectors of ∠BAC and ∠ABC respectively. In triangle ABI, the exterior angle ∠BIP = ∠IAB + ∠IBA = \(\frac{1}{2}\angle\)BAC + \(\frac{1}{2}\angle\)ABC = \(\frac{1}{2}[\angle\)BAC + ∠ABC] Again, ∠IBP = ∠IBC + ∠CBP = \(\frac{1}{2}\angle\)ABC + ∠PAC (∵ both are angles subtended by arc PC on the circle) = \(\frac{1}{2}\angle\)ABC + \(\frac{1}{2}\angle\)BAC ∴ In triangle IBP, ∠BIP = ∠IBP ∴ PI = PB = 15 cm