Question

the circles with centers P and Q touch each other at R A line passing through R meets the circles at A and B respectively. prove that seg AP parallel seg BQ

Answer 1

Angle ARP = Angle BRQ (Vertically opposite angles)

AP and PR are radii of the first circle, so they are equal

Triangle APR is isosceles, and so Angle PAR = Angle ARP


BQ and QR are radii of the second circle, so they are equal

Triangle BQR is isosceles, and so Angle QBR = Angle QRB

It follows that < PAR = < QBR 

.....and seg AP parallel seg BQ with AB being the transversal

Answer 2

Answer:

Step-by-step explanation:

Angle ARP = Angle BRQ (Vertically opposite angles)

AP and PR are radii of the first circle, so they are equal

Triangle APR is isosceles, and so Angle PAR = Angle ARP

BQ and QR are radii of the second circle, so they are equal

Triangle BQR is isosceles, and so Angle QBR = Angle QRB

It follows that < PAR = < QBR

.....and seg AP parallel seg BQ with AB being the transversal