Question

Two circles with centre o and p intersecrt at a and d if aob and a pc are diameter.proove that b c and d are collinear

Answer 1

Answer:

GIVEN: Circles with centre O & O' , AC is tangent to the circle with centre o' at A. OO' meets AB at D.

TO PROVE THAT: angle BAO = angle CAO PROOF: In triangle O'AO

O'A = O'O ( being radii of the same circle with centre O')

So, angle O'AO = angle O'OA = y……..(1)

O'A is perpendicular to tangent AC ( as theorem states that tangent to any circle at any point is perpendicular to the radius through that point.)

So, Angle O’AC = 90°…………..(2)

Eq(2) - Eq(1)

Angle O'AC — AngleO'AO

= Angle OAC = 90°— y ………..(3)

Now since ,

O is equidistant from A& B ( being radii of the same circle with centre O)

That implies that O lies on the perpendicular bisector of the segment AB.

Similarly, O' is equidistant from A & B( being radii of the same circle with centre O')

That implies that O' lies on the perpendicular bisector of the segment AB.

Hence conclude that OO' is perpendicular bisector of AB , at D ( by theorem )

Now, in Triangle ADO

Angle ADO = 90° ( proved above)

Angle DOA = y ( by eq (1) )

Therefore, third angle OAD =90° — y………(4)

By comparing eq(3) & (4)

We have, angle OAC = angle OAD

[Hence proved]