In a Geometry two column proof Prove: m( ∠ ABC) = 1/2 [ m(*arc*ADC) – m(*arc*AEC)] Look at picture below?

Question

Given: Circle O with tangent lines AB and CB and secant BD

Hint: 
POSTULATE 7.3= If A, B, and C are three points on the same arc, and B is between A and C then;
m(*arc*AC) = m(*arc*AB) + m(*arc*BC)
m(*arc*BC) = m(*arc*AC) - m(*arc*AB)
m(*arc*AB) = mm(*arc*AC) - m(*arc*BC)

THEOREM 7.7= The measure of an angle formed by the intersection of a tangent and a secant outside a circle is equal to one-half the difference of the measures of the intercepted arcs

Hosam · Accepted Answer

A sketch of the proof,  

Connect OB, OA, and OC, triangles OAB, and OCB are right triangles, so

  angle ABC = 180 - angle AOC

Angle AOC is a m(*arc* AEC), therefore,

  angle ABC = 180 - m(*arc* AEC) = 1/2 ( 360 - 2 m(*arc* AEC) )

                  = 1/2 ( 360 - m(*arc* AEC) - m(*arc* AEC) )

but 360 - m(*arc* AEC) = m(*arc* ADC), hence

   angle ABC = 1/2 ( m(*arc* ADC) - m(*arc* AEC) )

In a Geometry two column proof Prove: m( ∠ ABC) = 1/2 [ m(arcADC) – m(arcAEC)] Look at picture below?

1 Answer