Question

in a circular group O is the centre ,AB the diameter,C and D are any two points on the circumference so that C is the mid - point of arc BD prove that triangle AOC and triangle COA are equal in area​

Answer 1

Explanation:

1st method:-

Using congurency:-

In ∆AOC AND ∆COD

1) CO= CO {common side}

2) <OAC = <ODC { inscribed angel on same base CO}

3)OD = OC {radius}

4)∆AOC ≅ ∆COD {SAS fact}

5)∆AOC = ∆ COD {area of congurent traingle is equal}



2nd method:-

Using property of traingles:-

Construction:- Join DA and CB

<ACB = 90° { angle substended by diameter}

CO⊥AB

Now we know that,

Area of traingle = 1/2 B*H

CO is height { of both traingle's}

AO is base {of both traingle's}

Now, ∆AOC= 1/2*CO*AO--1

and, ∆ COD = 1/2 *CO*AO--2

∆AOC= ∆COD {from above 1 and 2}