TA, TB ARE TANGENT SEGMENTS TO A CIRCLE(CENTRE O) FROM ONE EXTERNAL POINT T AND OT INTERSECTS THE CIRCLE IN P. PROVE THAT AP BISECTS THE ANGLE TAB
Draw AO, BO, AB and BP.......label the intersection of TO and AB as point F
With the above additions, I believe the image below is what you describe :
By SSS, triangle ATO congruent to triangle BTO.....thus angle ATO = angle BTO
Thus, by SAS, triangle ATF congruent to triangle BTF
Thus, FA = FB
And a radius bisecting a chord does so at right angles... so, OP is perpendicular to AB
Thus, by LL, triangle PAF congruent to triangle PBF
Thus angle PAF = angle PBF
And, by Euclid, m< TAP = 1/2 minor arc PA
But, angle PBF = angle PBA .....and PBA is an inscribed angle = 1/2 minor arc PA
But, it was shown that angle PBF = angle PAF......thus, angle PAF also = 1/2 minor arc PA
So.......m<TAP = m<PAF = m<PAB
So....m<TAP = m<PAB, and AP bisects TAB
Draw AO, BO, AB and BP.......label the intersection of TO and AB as point F
With the above additions, I believe the image below is what you describe :
By SSS, triangle ATO congruent to triangle BTO.....thus angle ATO = angle BTO
Thus, by SAS, triangle ATF congruent to triangle BTF
Thus, FA = FB
And a radius bisecting a chord does so at right angles... so, OP is perpendicular to AB
Thus, by LL, triangle PAF congruent to triangle PBF
Thus angle PAF = angle PBF
And, by Euclid, m< TAP = 1/2 minor arc PA
But, angle PBF = angle PBA .....and PBA is an inscribed angle = 1/2 minor arc PA
But, it was shown that angle PBF = angle PAF......thus, angle PAF also = 1/2 minor arc PA
So.......m<TAP = m<PAF = m<PAB
So....m<TAP = m<PAB, and AP bisects TAB
As an addendum to this problem......let me prove the assertion concerning Euclid....namely that m< TAP = 1/2 minor arc PA
Draw diameter AB
And m< AFB = m< TAP + m< PAB
And m< AFB = 1/2 arc APB
And m< PAB = 1/2 minor arc PB
Thus :
m< AFB - m<PAB = m< TAP and, by substitution
1/2arc APB - 1/2 minor arc PB = m<TAP
1/2 [ arc APB - minor arc PB] = m< TAP
1/2 [ minor arc PA] = m< TAP