Question
In figure 3.97, circles with centres C and D touch internally at point E. D lies on the inner circle. Chord EB of the outer circle intersects inner circle at point A. Prove that, seg EA ≅seg AB.
✓
Answer
We see that the line joining D to E passes through C.
In the smaller circle,
A lies in the semicircle,
∴ ∠EAD = 90°
⇒ DA is perpendicular on the chord EB of the bigger circle.
We know that perpendicular from the center bisects the chord.
Therefore, EA = AB.
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