Question

O is the centre of the circle in which ab and ac are congruent chords radius open is perpendicular to chord ab and oq is perpendicular to chord ac if angle pba is 30 degree then prove that seg pb is parallel to seg qc

Answer 1

Answer:

Given, O is the center of a circle.

AB=AC

OP⊥AB

OQ⊥AC

∠PBA=30

∘

To prove :- BP∥QC

Construction : Join BC, OC and OB, as shown in the figure above.

Proof : AB=AC

We know that, angle opposite to equal sides of a triangle are equal.

∴ ∠ACB=∠ABC __ (1)

OC=OB [∵ radius]

∠OCB=∠OBC __ (2)

Using (1), we have:

∠ACB=∠ABC

⇒∠ACO+∠OCB=∠ABO+∠OBC

Using (2), we can say:

⇒∠ACO=∠ABO __ (3)

In △OXC and △OYB

∠OXC=∠OYB=90

o

[∵OQ⊥AC&OP⊥AB]

∠AOC=∠ABO [using (3)]

OC=OB [Radii]

∴△OXC≅△OYB [AAS]

∠QOC=∠POB [CPCT] __ (4)

Now in △QOC and △POB,

OQ=OB[∵ radius]

∠QOC=∠POB (using (4))

OC=OP(∵ radius)

∴△QOC≅△POB [SAS]

∴OQC=∠OBP [CPCT]

These are alternate interior angles formed between QC and PB.

Hence, QC∥PB .