Question

If A , B , C and D are angles of a cyclic quadrilateral prove that $\cos A + \cos B + \cos C + \cos D = 0$.

Answer 1

Hint: As A , B , C and D are the angles of a cyclic quadrilateral , by the theorem opposite angles of a cyclic quadrilateral are supplementary. We get $C = {180^ \circ } - A,D = {180^ \circ } - B$and using this in our given equation we get the required proof.

Step by step solution :
We are given a cyclic quadrilateral
And we are given that A , B , C and D are its angles
We know that the opposite angles of a cyclic quadrilateral are supplementary
That is , the opposite angles sum upto to 180 degrees
Here we can see , A and C are opposite angles and B and D are opposite angles
So ,
$
   \Rightarrow A + C = {180^ \circ } \\
   \Rightarrow B + D = {180^ \circ } \\
 $
Now we can rewrite these equations as
$
   \Rightarrow C = {180^ \circ } - A \\
   \Rightarrow D = {180^ \circ } - B \\
 $
Now we need to find the value of $\cos A + \cos B + \cos C + \cos D$
Using the values of C and D from the above equations we get
$ \Rightarrow \cos A + \cos B + \cos \left( {180 - A} \right) + \cos \left( {180 - B} \right)$
Since $\cos \left( {180 - \theta } \right) = - \cos \theta $ , we get
$ \Rightarrow \cos A + \cos B - \cos A - \cos B = 0$
Hence proved

Note :
1) A cyclic quadrilateral or inscribed quadrilateral is a quadrilateral whose vertices all lie on a single circle. This circle is called the circumcircle or circumscribed circle, and the vertices are said to be concyclic.
2) The product of the diagonals of a quadrilateral inscribed in a circle is equal to the sum of the product of its two pairs of opposite sides.
3) The product of the diagonals of a quadrilateral inscribed in a circle is equal to the sum of the product of its two pairs of opposite sides.
4) The perpendicular bisectors of the fours sides of the inscribed quadrilateral intersect at the center o.