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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

aditya@calc.com Jan 10, 2016

#1**+15 **

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

CPhill Jan 10, 2016

#1**+15 **

Best Answer

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

CPhill Jan 10, 2016

#2**+10 **

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

CPhill Jan 11, 2016