Question

The construction of a tangent to a circle given a point outside the circle can be justified using the second corollary to the inscribed angle theorem. An alternative proof of this construction is shown below. Complete the proof. (5 points)

Given: Circle C is constructed so that CD = DE = AD; is a radius of circle C.

Prove: is tangent to circle C.

Answer 1

I got a good grade on it yw ;)

Answer 2

Proof:

Draw circle C with center at point A and radius AD = CD = DE.

Draw point P outside the circle C.

Draw segment AP and extend it to intersect the circle at point B.

Draw segment BD.

Draw segment CP.

Note that triangle BCD is isosceles, since CD = BD. Therefore, angle BDC = angle CBD.

Since angle BDC is an inscribed angle that intercepts arc BC, and angle CBD is an angle that intercepts the same arc, then angle BDC = angle CBD = 1/2(arc BC).

Since CD = DE, then angle CED = angle CDE. Therefore, angle DCE = 1/2(arc BC).

Since angles BDC and DCE are equal, then angles BDC and CBD are also equal, and triangle BPC is isosceles. Therefore, segment BP = segment PC.

Since BP = PC, then segment CP is perpendicular to segment BD, by the Converse of the Perpendicular Bisector Theorem.

Therefore, segment CP is a tangent to circle C at point B.

Hence, the proof is complete.