Question

Points S, O, M, and I are concyclic such that arc SO=arc IM. If the chord SM=IO are intersected at the points K. Prove that area of triangle SOK=area of triangle IMK and SM=IO

Answer 1

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:

Area of a Triangle:

Formula to find the area of a triangle is

• 
  `A=(1)/(2)*(Base)(Height)`

Base can be any side of a triangle but height must be the side perpendicular to the base

Concyclic points:

Points which lie on the same circle having the same distance from the center of the circle.

Given Data:

Arc SO= Arc IM

chord SM= chord IO

Proof of Area of Triangle SOK=Area of Triangle IMK:

As the given points are concyclic points, so

• SK=OK

• KM=IM

Any of the above point is radius of the circle.

Thus,

• SK=OK=KM=IM

Are of Triangle SOK:

• 
  `Area\ of\ the\ Triangle\ SOK = (1)/(2)*(SK)(OK)`

where, SK is the Base for triangle SOK and OK is the Height for the triangle SOK

Area of Triangle IMK:

• 
  `Area\ of\ Triangle\ IMK= (1)/(2)*(KM)(IK) \\`

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\ the\ Triangle\ SOK = (1)/(2)*(SK)(OK)` =
`Area\ of\ Triangle\ IMK= (1)/(2)*(KM)(IK) \\`

OR

`(1)/(2)*(SK)(SK)`
`=(1)/(2)*(SK)(SK)`

thus proved

Proof of SM=IO:

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