Question

ABCD is a quadrilateral. If $$\mathrm{\angle }A=\mathrm{\angle }BCE$$, is the quadrilateral a cyclic quadrilateral ? Give reasons.

Answer 1

Hint: To prove the above statement use the property of cyclic quadrilateral i. e. opposite angles of a cyclic quadrilateral are always Supplementary i.e. their summation is $180{}^{\circ }$.
Complete step-by-step answer:
To prove that quadrilateral ABCD is a cyclic quadrilateral, we should know the necessary and sufficient condition for a quadrilateral to be cyclic which is given below,
Concept: Opposite angles of a cyclic quadrilateral are always Supplementary i.e. their summation is $180{}^{\circ }$.
Which means we have to prove,
$$\mathrm{\angle }A+\mathrm{\angle }DCB=180{}^{\circ }and\mathrm{\angle }ADC+\mathrm{\angle }B=180{}^{\circ }$$
Let,
$$\mathrm{\angle }A=\mathrm{\angle }BCE=\theta$$……………………………….. (1)
From figure we can easily see that $$\mathrm{\angle }DCE$$ is a straight angle,
$$\therefore \mathrm{\angle }DCE=180{}^{\circ }$$
But, from figure we can also write,
$$\mathrm{\angle }DCE=\mathrm{\angle }DCB+\mathrm{\angle }ECB$$
Therefore from above two equations we can write,
$$\mathrm{\angle }DCB+\mathrm{\angle }ECB=180{}^{\circ }$$

Put the value of equation (1),
$$\therefore \mathrm{\angle }DCB+\theta =180{}^{\circ }$$
$$\therefore \mathrm{\angle }DCB=180{}^{\circ }-\theta$$…………………………………… (2)
Now we will add $$\mathrm{\angle }Aand\mathrm{\angle }DCB$$ to see if they are supplementary.
$$\mathrm{\angle }A+\mathrm{\angle }DCB=\theta +(180{}^{\circ }-\theta )$$ [From (1) and (2)]
By giving separate signs in bracket we can write,
$$\therefore \mathrm{\angle }A+\mathrm{\angle }DCB=\theta +180{}^{\circ }-\theta$$
$$\therefore \mathrm{\angle }A+\mathrm{\angle }DCB=180{}^{\circ }$$……………………………………… (3)
Now, as we all know summation of all angles of a square is always $$360{}^{\circ }$$,
$$\therefore \mathrm{\angle }A+\mathrm{\angle }B+\mathrm{\angle }DCB+\mathrm{\angle }ADC=360{}^{\circ }$$
By rearranging above equation,
$$\therefore \mathrm{\angle }A+\mathrm{\angle }DCB+\mathrm{\angle }B+\mathrm{\angle }ADC=360{}^{\circ }$$
Put the value of Equation (3) in above equation,
$$\therefore 180{}^{\circ }+\mathrm{\angle }B+\mathrm{\angle }ADC=360{}^{\circ }$$
$$\therefore \mathrm{\angle }B+\mathrm{\angle }ADC=360{}^{\circ }-180{}^{\circ }$$
$$\therefore \mathrm{\angle }B+\mathrm{\angle }ADC=180{}^{\circ }$$……………………………………….. (4)
From equations (3) and (4),
We can conclude that $$\mathrm{\angle }A,\mathrm{\angle }DCB$$ and $$\mathrm{\angle }ADC,\mathrm{\angle }B$$ are pairs of supplementary angles and from the concept we have discussed earlier we can conclude that $$◻ABCD$$ is a cyclic quadrilateral.
Hence Proved.
Note:
1) The concept given by “opposite angles of a cyclic quadrilateral are always Supplementary” is very much essential to prove any quadrilateral as a cyclic quadrilateral.
2) I have inscribed above the quadrilateral in a circle geometrically therefore it is proved experimentally also. I have taken the mentioned angle as $$60{}^{\circ }$$.