Question

Solve this question .​

Answer 1

We have  R^{2} > CP . CB. On the other hand, the power of P with respect to

the circumcircle of ABC is BP . PC = R^{2}  

OP2 . From these two equations we find that

OP2 = R2  

BP . PC > PC . CB  - BP . PC = PC^{2}  ,

from which OP > PC. Therefore, as in the first solution, we conclude that αβ < 90  .

Answer 2

Question :

Let ABC be an  acute-angled triangle with circumcentre O.  Let P on BC be the foot of the altitude  from A. Suppose that ∠BCA ≥ ∠ABC +  30°. Prove that ∠CAB + ∠COP < 90° .

Answer:

Refer the attachment for the figure of the question .

In Δ OPC ,

OP > PC

⇒ ∠OCP > ∠COP

Angles which are longer have longer opposite sides .

Now in Δ OCB ,

∠OCB + ∠OBC + ∠COB = 180° [ By angle sum property ]

∠ OCB = ∠ OBC [ Radii are equal and isosceles triangles have equal base angles ]

⇒ ∠COB + 2∠OCB = 180°

∠OCB is the same as ∠OCP and hence :

∠ COB + 2∠OCP = 180°

Now since ∠COP < ∠OCP :

⇒ ∠COB + 2∠COP < 180°

⇒ ∠COB/2 + ∠COP < 90°

We know that angle subtended at the centre is twice the angle subtended at the periphery .

Hence ∠CAB = ∠COB/2

⇒ ∠CAB + ∠COP < 90°

Hence proved .