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Question
In the given figure, P is the centre of the circle. chord AB and chord CD intersect on the diameter at the point E. If ∠AEP ≅ ∠DEP then prove that AB = CD.
Solution
Given: P is the centre of the circle.
Chord AB and chord CD intersect on the diameter at the point E.
∠AEP ≅ ∠DEP
To prove: AB = CD
Construction: Draw seg PM ⊥ chord AB such that A-M-B and draw seg PN ⊥ chord CD such that C-N-D
Proof:
∠AEP ≅ ∠DEP ...(Given)
∴ Seg ES is the bisector of ∠AED.
Point P is on the bisector of ∠AED.
∴ PM = PN ...(Every point on the bisector of an angle is equidistant from the sides of the angle.)
∴ chord AB ≅ chord CD ...(Chords equidistant from the center of the circle are congruent.)
∴ AB = CD ...(Length of congruent segments)
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