Answer:
Step-by-step explanation:
From the figure attached,
CP is an angle bisector of angle BCD and BP is the angle bisector of angle CBE
Therefore, m∠DCP ≅ m∠BCP
and m∠PBE ≅ m∠PBC
m∠A + m∠CBA = m∠BCD
m∠A + (180° - m∠CBE) = m∠BCD
m∠A + 180° = m∠CBE + m∠BCD
m∠A + 180° = 2(m∠PCB) + 2(m∠PBC) [Since m∠CBE = 2m∠PCB and m∠BCD = 2(m∠PBC)
m∠A + 180° = 2(m∠PCB + m∠PBC)
m∠A + 180° = 2(180° - m∠BPC) [Since m∠PCB + m∠PBC + m∠BPC = 180°]
[tex][\frac{1}{2}(m\angle A)]+90=180 - m\angle BPC[/tex]
m∠BPC = 180 - [tex][\frac{1}{2}(m\angle A)]-90[/tex]
m∠BPC = 90 - [tex]\frac{1}{2}m\angle A[/tex]
