Two circles touch internally at A. The tangent at P on the smaller circle cuts the larger circle at Q and R. Prove that AP bisects (Angle)RAQ.
Thanks in advance.
Edit: This question is question 28 on page 18 of New Senior Mathematics, 3U by Fitzpatrick
Unfortunately I don't know how to do a diagram ....... so I'll just have to describe.
Draw the 2 internally touching circles. Choose any reasonable point P on the smaller circle and draw the tangent RPQ. Draw tangent at A and let this intersect QPR produced at T. Draw the lines AQ and let this intersect smaller circle at S........ Draw PA and AR.
Let angle SPQ = @ and angle RAT = & and angle APR = $.
.: by alternate segment thm, angle SAP = @ and angle AQR = &. For triangle APQ,
angle APR = $ = angle QAP + angle AQP
(ext angle of triangle = sum of int opp angles)
i.e. $ = @ + &
Since TA and TP are tangents from an exterior point T to smaller circle TA = TP
.: angle TAP = angle TPA (= $)
(base angles of isosceles triangle)
.: angle TAR
( = &) + angle RAP = $ = @ + &
.: angle RAP = @
.: angle RAP = angle PAQ
(= @)
i.e AP bisects angle RAQ
QED
You can replace the 3 angles @ & and $ with some suitable Greek letters like theta, phi and psi to make it easier to follow the proof; don't know how to access these Greek letters.
