Question

Two cords AB and cd intersect at point p in the circle. Prove that the sum of the angles distended by the arcs AC and BD as the centre O is equal to Twice of the angle APC

Answer 1

Step-by-step explanation:

as we know that the angle subtended by its chord is twice the angles create by the rest of its part or centre.

so with this statement we can epress that

∠AOD = 2∠BDC__________(1)

and

∠BOC = 2∠BDC_____________(2)

Add (1)and(2)

∠AOD + ∠BOC = 2 ∠ABD + 2 ∠BDC

∠AOC + ∠BOC = 2 (∠ABD+∠BDC)___________(3)     (from 1 and 2)

now,

∠APD =  ∠PBD + ∠PDB ___________(4)

now accordingto the figure

∠PBD = ∠ABD and

∠PDB = ∠CDB

so by substituting the values of ∠ABD and ∠BDC in equation 3 we get

∠AOD +∠BOC = 2(∠ABD+∠BDC)

∠AOD +∠BOC = 2 (∠PBD+∠PDB)

∠AOD +∠BOC = 2∠APD

Hence Proved!