Problem: ABCD is a rectangle (see the accompanying diagram) with P any point on AB. PS⊥BD and PR⊥AC. AF⊥BD and PQ⊥AF. Then PR+PS is equal to:
Answer Choices:
A. PQ
B. AE
C. PT+AT
D. AF
E. EF
Solution:
△PTR∼△ATQ;AQPR=ATPT
PT=AT(∠PAT=∠PBS=∠APT)
PR=AQ;PS=QF
PR+PS=AQ+QF=AF
or
∠SBP=∠TPA=∠TAP
A,Q,R,P are concyclic.
arcPR=arcAQ
PR=AQ
PS=QF;PR+RS=AF
