Geometry-P3

Problem↑

Solution (Several ways):

Make CE⊥AP, Connect BE.

Since m∠CEP=90° and m∠EPC=60

m∠ECP=30°

According to the 30-60-90 triangle theorem

2EP=PC

since 2BP=PC

EP=BP

m∠EBP=m∠BEP (EP's oppsite angle is congruent to BP's)

since m∠APC=60

m∠APB=120

m∠EBP=m∠BEP=(180-120)/2=30

Since m∠EBP=m∠ECP

BE=EC

m∠EBP=45-30=15

m∠BAP=60-45=15 (exterior angles theorem)

Since m∠EBP=m∠BAP

BE=AE=EC

Since AE=EC and ∠AEC=90

∠EAC=∠ACE=45

Since m∠ACE+m∠ECP=m∠C (angles addition postulate)

m∠C=45+30=75°