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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

Best Answer 

 #1
avatar+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

 

 

cool cool cool

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: 

2+0 Answers

 #1
avatar+78577 
+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

 

 

cool cool cool

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
avatar+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

 

 

cool cool cool

CPhill  Jan 11, 2016

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