SOLUTION: AB is a chord of a segment of a circle and C is any point on the arc of the segment .Prove that m<CAB+m<CBA is the same for every position of C.

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Question 1068287: AB is a chord of a segment of a circle and C is any point on the arc of the segment .Prove that m
Found 3 solutions by ikleyn, Fombitz, KMST:
Answer by ikleyn(52858)   (Show Source): You can put this solution on YOUR website!
.
m < CAB  +m < CBA  = 180 - m < c.


m < C is the same for all positions of the point C in the circle and is half of the measure of the arc AB, 
      since the angle C is an inscribed angle.


It implies that  m < CAB  +m < CBA  = const,

QED.

On inscribed angles see the lesson
    - An inscribed angle in a circle,
in this site.


Also, you have this free of charge online textbook on Geometry
    - GEOMETRY - YOUR ONLINE TEXTBOOK
in this site.

The referred lesson is the part of this textbook under the topic
"Properties of circles, inscribed angles, chords, secants and tangents".



Answer by Fombitz(32388)   (Show Source): You can put this solution on YOUR website!
Draw a line that is the perpendicular bisector of AB and find the point O that is the intersection of the bisector and the circle. Label that point O.
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OACB forms a cyclic quadrilateral and the opposite angles are supplementary.
So,

However AOB is fixed and independent of the choice of C.
That means that ACB is also fixed.
You also know that,


Since the right hand side is a fixed value, the sum of the other two angle is also fixed.
Answer by KMST(5328)   (Show Source): You can put this solution on YOUR website!
Angles CAB and CBA are inscribed angles
that intercept arcs CA and CB.
Each of those angles measures half of
the measure of the intercept arc.
The sum of the measures of those angles
is half the sum of the measures
of the intercepted arcs CA and CB.
The sum of the measures of those arcs
Is the measure of arc AB,
regardless of the position of point C.
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