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ABCD is a cyclic quadrilateral whose diagonals intersect at O. If $\angle ACB=50{}^\circ$ and $\angle ABC=110{}^\circ$, find $\angle BDC$.


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Hint: We can solve this problem by using the property of angle inscribed by the arc, i. e. Angles inscribed by the same arc on the circumference of a circle are always EQUAL, and the property of a triangle.


Complete step-by-step answer:
Firstly, we will write the given values,
$\angle ACB=50{}^\circ$ and $\angle ABC=110{}^\circ $……………………………….. (1)
To find the required angle we should know the key concept which is given below,
Angles inscribed by the same arc on the circumference of a circle are always EQUAL, which means, angles made by arc BC at point D and A on the circumference of the circle are equal.
\[\therefore \angle BDC=\angle BAC\]……………………………………………….. (2)
To find the value of \[\angle BAC\] consider\[\vartriangle ABC\],

As the sum of angles of a triangle is always \[180{}^\circ \]
\[\therefore \angle ABC+\angle BAC+\angle ACB=180{}^\circ \]
Put the values of (1) in the above equation,
\[\Rightarrow 110{}^\circ +\angle BAC+50{}^\circ =180{}^\circ \]
\[\Rightarrow \angle BAC=180{}^\circ -160{}^\circ \]
\[\Rightarrow \angle BAC=20{}^\circ \]
Now put the value of equation (2) in the above equation,
\[\Rightarrow \angle BDC=\angle BAC=20{}^\circ \]
\[\therefore \angle BDC=20{}^\circ \]
Therefore the value of \[\angle BDC\] is \[20{}^\circ \].

Note: In the problem they have just mentioned that diagonals are passing through point O, but it is not clear that O is the centre of the circle. If we consider YOU as center then, \[\angle DCA\] will become \[40{}^\circ \] and we will get the final answer as \[\angle BDC=50{}^\circ \] which is a wrong answer.
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