Question

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Answer 1

$\underline{\mathfrak{\huge{Question:}}}$

In the figure, AB is a diameter of the circle, CD is a chord equal to the radius of the circle. AC and BD when extended intersect at a point E. Prove that /_AEB = 60°.

The figure has been attached in the answer.

$\underline{\mathfrak{\huge{Answer:}}}$

$\underline{\sf{Construction:}}$ Join OD, DA, and OC

Triangle COD proves to be an equilateral triangle. As :-

OC = OD = CD ( CD = radius length, Given ; the other two are the radius of the same circle )

/_COD = 60° ( Angles of an equilateral triangle are equal to 60° each )

2/_CAD = /_COD ( Chord CD subtends angles both, the centre as well as the circumference of the circle. The angle subtended by a chord on the centre is double of the angle subtended by it on any other point of the same circle )

/_CAD = $\frac{60}{2}$ ( Since /_COD = 60° )

=》 /_CAD = 30°

2/_ADB = /_AOB = $\frac{180}{2}$ = 90° ( The angle subtended by a chord on the centre is double of the angle subtended by it on any other point of the same circle )

$\textbf{In Triangle ADE:}$

/_ADB = /_CAD + /_AED ( exterior angle property)

=》 90° = 30° + /_AED

=》 /_AED = 90° - 30°

=》 /_AED = 60°

$\underline{\tt{Hence\:Proved!!}}$