Please find the file attached below (figure for the question)
To prove that area of the two triangles is equal we must know the formula for the area of a triangle and concept of concyclic points as given below:
<u>Area of a Triangle:</u>
Formula to find the area of a triangle is
Base can be any side of a triangle but height must be the side perpendicular to the base
<u>Concyclic points:</u>
Points which lie on the same circle having the same distance from the center of the circle.
<u>Given Data</u>:
Arc SO= Arc IM
chord SM= chord IO
<u> </u><u>Proof of Area of Triangle SOK=Area of Triangle IMK:</u>
As the given points are concyclic points, so
Any of the above point is radius of the circle.
Thus,
<u>Are of Triangle SOK:</u>
where, SK is the Base for triangle SOK and OK is the Height for the triangle SOK
<u>Area of Triangle IMK:</u>
- <u></u>
<u></u>
Where, KM is Base of the triangle IMK and IK is Height of the triangle IMK
As we know
SK=OK=KM=IM
We can say directly that area of both the triangles is same
=![Area\ of\ Triangle\ IMK= \frac{1}{2}*(KM)(IK) \\](https://tex.z-dn.net/?f=Area%5C%20of%5C%20Triangle%5C%20IMK%3D%20%5Cfrac%7B1%7D%7B2%7D%2A%28KM%29%28IK%29%20%5C%5C)
OR
![\frac{1}{2}*(SK)(SK)](https://tex.z-dn.net/?f=%5Cfrac%7B1%7D%7B2%7D%2A%28SK%29%28SK%29)
![=\frac{1}{2}*(SK)(SK)](https://tex.z-dn.net/?f=%3D%5Cfrac%7B1%7D%7B2%7D%2A%28SK%29%28SK%29)
thus proved
<u>Proof of SM=IO:</u>
As points are concyclic so they all have same distance from the center of the circle
i.e.SK=OK=KM=IM
thus SM=IO