$\angle IQP=\angle IPQ = \frac{\angle A}{2}$ $\angle CQC_1=360^o-2\angle CQP =360^o-2( 90^o+\angle IQP)=180^o-\angle A= \angle QIP$ So $CQ=C_1Q=QI=PI \to CC_1=QP$ $\triangle CQP= \triangle RCC_1$ because $CQ=RC, QP=CC_1, \angle RCC_1=90^o+\angle QCC_1=90^o+\angle IQP= \angle CQP$ So $RC_1=CP=C_1P$ So $\triangle RIC_1 = \triangle PIC_1$ and so $C_1$ lies on $BI$ - bisector of $\angle B$ Same way $C_2$ is on bisector $AI$ so $\angle C_1IC_2 = \angle BIC=180^o-\frac{\angle A+\angle B}{2}=135^o$