ABC is a cyclic equailateral triangle. If D be any point on the circular are BC on the oppsite side of the point A, then prove that `DA=DB+DC`

Explanation

Let ABC be a cyclic equailateral angle inside the circle with centre at O,D is any point on the circular are BC on the opposite side of the point A.
We have to prove that `DA=DB+DC`

Construction Let us cut a part De from DA equal to DC. i.e, DC= DE and let us join C,E
In `Delta CDE, DC= DE ` [ as per construction]
`:. angle DCE=angle DEC.....(1)`
`:. Delta ABC` is an equilateral triangle,
`angleBAC=angle ACB=angle CBA=60^(@) and AB BC= CA ......(2)`
Now, two angles in circle produced by the are AC are `angle ADC and angle ABC`
`:. angle ADC= 60^(@)`[ by (2)]
`:. angle CDE= 60^(@)......(3)`
`:."in" Delta CDE, angle CDE+ angle DCE= 180^(@)+ angle DEC=180^(@)`
or, `60^(@)+angleDCE+ angle DCE= 180^(@)`[ from (1) and (3)]
or, ` 2angle DCE =180^(@)-60^(@) or, angle DCE=(120^(@))/(2)=60^(@)`
`:. angle DEC=60^(@)` i.,e, each and very angle of `Delta CDE` is `60^(@), Delta CDE` is equilateral.
`:. CD=DE=CE....(4)`
Now, `angle ACE+angleBCE= angleACB=60^(@)= angle DCE= angle BCD+angle BCE`
`:. angle ACE= angle BCD........(5)`
Again in `Delta`'s ACE and `Delta BCD`
`angle CAE= angle CBD[ :.` bothare angles in cricle produced by the same chord CD]
`:. angle ACE= angle BCD` [by (5)] and AC=BC [ `:. Delta ABC` is equilateral]
`:. Delta ACE~= BCD [ :.` by the A-A-S condition of congruency]
`:. AE=BD [ :.` similar sides of congruent tirangles]........(6)
Therefore, `DA=DE+AE=DC+BD [ :. DE= DC and AE= BD]`
Hence `DA=DB+DC`