F.5 m Tangent to Circles 快40分!

(a) DQ=DA=a and CQ=CB=b
DC=DQ+QC=a+b


(b) AD | AB, PQ | AB
AD//RQ
△ACD~△RCQ
QR/DA=QC/DC
QR/a=b/(a+b)
QR=ab/(a+b)


(c) △ACD~△RCQ
RC/AC=QC/DC
RC/AC=b/(a+b)
1-RC/AC=1-b/(a+b)
(AC-RC)/AC=[(a+b)-b]/(a+b)
AR/AC=a/(a+b)

rt∠△APR~rt∠△ABC
PR/BC=AR/AC
PR/b=a/(a+b)
PR=ab/(a+b)

∴ PR=QR