+0

# solve this

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+128

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
+78577
+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
edited by CPhill  Jan 10, 2016
edited by CPhill  Jan 10, 2016
edited by CPhill  Jan 10, 2016
edited by CPhill  Jan 11, 2016
edited by CPhill  Jan 11, 2016
edited by CPhill  Jan 11, 2016
Sort:

#1
+78577
+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
edited by CPhill  Jan 10, 2016
edited by CPhill  Jan 10, 2016
edited by CPhill  Jan 10, 2016
edited by CPhill  Jan 11, 2016
edited by CPhill  Jan 11, 2016
edited by CPhill  Jan 11, 2016
#2
+78577
+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

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