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Answer :

As, Angles in the same segment are equal, we get,

\(\angle{BDC}\) = \(\angle{BAC}\)

\(\therefore \) \(\angle{BDC}\) = \(30^\circ\).

In \(\triangle{DBC}\), we have,

\(\angle{BDC}\) + \(\angle{DBC}\) + \(\angle{BCD}\) = \(180^\circ\)

\(\Rightarrow \) \(30^\circ\) + \(70^\circ\) + \(\angle{BCD}\) = \(180^\circ\)

\(\Rightarrow \) \(100^\circ\) + \(\angle{BCD}\) = \(180^\circ\)

\(\Rightarrow \) \(\angle{BCD}\) = \(180^\circ\) - \(100^\circ\)

\(\therefore \) \(\angle{BCD}\) = \(80^\circ\)

If AB= BC, \(\angle{BCA}\) = \(\angle{BAC}\) = \(30^\circ\)

(Since, Angles opposite to equal sides in a traingle are equal)

Now,

\(\angle{ECD}\) = \(\angle{BCD}\) - \(\angle{BCA}\) = \(80^\circ\) - \(30^\circ\)

(Given \(\angle{BCD}\) = \(80^\circ\) and \(\angle{BCA}\) = \(30^\circ\))

\(\Rightarrow \) \(\angle{ECD}\) = \(30^\circ\)

\(\therefore \) \(\angle{EDC}\) = \(30^\circ\)

- In figure A, B and C are three points on a circle with centre O such that \(\angle{BOC}\) = \(30^\circ\) and \(\angle{AOB}\) = \(60^\circ\). If D is a point on the circle other than the arc ABC, find \(\angle{ADC}\).Let PQ = QR = PR = X
- A chord of a circle is equal to the radius of the circle. Find the angle subtended by the chord at a point on the minor arc and also at a point on the major arc.
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- In figure, A, B and C are four points on a circle. AC and BD intersect at a point E such that \(\angle{BEC}\) = \(130^\circ\) and \(\angle{ECD}\) = \(20^\circ\). Find \(\angle{BAC}\).
- If diagonals of a cyclic quadrilateral are diameters of the circle through the vertices of the quadrilateral, prove that it is a rectangle.
- If the non-parallel sides of a trapezium are equal, prove that it is cyclic.
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