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Answer :

In \(\triangle{ABC}\),

\(\angle{AED}\) = \(180^\circ\) -\(130^\circ\) = \(50^\circ\)

(linear pair of angles)

\(\therefore \) \(\angle{CED}\) = \(\angle{AED}\) = \(50^\circ\)

(Vertically opposite angles)

Also, \(\angle{ABD}\) = \(\angle{ACD}\)

(Since, the angles in the same segment are equal)

\(\Rightarrow \) \(\angle{ABE}\) = \(\angle{ECD}\)

\(\therefore \)
\(\angle{ABE}\) = \(180^\circ\) ...(ii)

Now, in \(\triangle{CDE}\),

\(\angle{BAC}\) + \(20^\circ\) + \(50^\circ\) = \(180^\circ\)

(from eq.(i) and (ii))

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

\(\Rightarrow \) \(\angle{BAC}\) = \(180^\circ\) - \(70^\circ\)

\(\therefore \) \(\angle{BAC}\) = \(110^\circ\)

- 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.
- In figure, \(\angle{PQR}\) = \(100^\circ\) ,where P, Q and R are points on a circle with centre O. Find \(\angle{OPR}\).
- In figure, \(\angle{ABC}\) = \(69^\circ\), \(\angle{ACB}\) = \(31^\circ\). Find \(\angle{BDC}\).
- ABCD is a cyclic quadrilateral whose diagonals intersect at a point E. If \(\angle{DBC}\) = \(70^\circ\), \(\angle{BAC}\) = \(30^\circ\), Find \(\angle{BCD}\). Further, if AB = BC, find \(\angle{EDC}\).
- 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.
- Two circles intersect at two points B and C. Through B, two line segments ABD and PBQ are drawn to intersect the circles at A, D and P, Q respectively (see figure). Prove that \(\angle{ACP}\) = \(\angle{QCD}\).
- If circles are drawn taking two sides of a triangle as diameters, prove that the point of intersection of these circles lie on the third side.
- ABC and ADC are two right angled triangles with common hypotenuse AC. Prove that \(\angle{CAD}\) = \(\angle{CBD}\).
- Prove that a cyclic parallelogram is a rectangle.

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