We will prove that, A tangent, DE, to a circle at A is parallel to a chord BC of the circle. Prove that A is equidistant from the extremities of the chord.
Solution:
Proof:
Statement |
Reason |
1. ∠DAB = ∠ACB |
1. Angle between tangent and chord is equal to the angle in the alternate segment. |
2. ∠DAB = ∠ABC |
2. Alternate angles and DE ∥ BC. |
3. ∠ACB = ∠ABC |
3. From statements 1 and 2. |
4. AB = AC ⟹ A is equidistant from B and C, the extremities of the chord. (Proved) |
4. From statement 3. |
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