In the adjoining figure AOB is a diameter,ARC DC =ARC BC,then prove that AD//OC
i)BOC=BC
ii)DAB=1/2BD=BC
iii)BOC=DAB
iv)AD//OC
I don't know,plz help me,its important
i)♢PAQ=♢SBR
ii)□PQRB-♢PAB=PQRB-♢SBR
iii)□AQRB=□PQRS
To prove that AD is parallel to OC, we need to show that the corresponding angles are equal.
Let's break down the steps for this proof:
1. Given: AOB is a diameter of the circle.
2. ARC DC = ARC BC (given)
3. OA = OB (a diameter bisects the circle and hence the arcs are equal)
4. Let's consider two cases:
Case 1: When A and C are on the same side of the diameter:
In triangle OAC,
angle AOC = angle OAC + angle OCA (angle sum property of triangles)
Since OA = OB (as they are radii of the same circle),
angle OAC = angle OCA (isosceles triangle property)
Therefore, angle AOC = angle OCA + angle OCA
But ARC DC = ARC BC (given)
So, angle OCA = angle OCB
Hence, angle AOC = angle OCB
Now, in triangle ADO, angle ADO = angle AOC (vertical angles are equal)
Therefore, angle ADO = angle OCB (substituting angle AOC = angle OCB)
Since angle ADO = angle OCB and angle AD and angle OC are corresponding angles, we can conclude that AD is parallel to OC.
Case 2: When A and C are on opposite sides of the diameter:
In this case, we can similarly prove that angle ADO = angle OCB, and thus AD is parallel to OC.
Therefore, in both cases, we have shown that AD is parallel to OC.
Hence, we have proved that AD is parallel to OC.