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Question

O is the centre of a circle in which AB and AC are congruent chords. Radius OP is perpendicular to chord AB and radius OQ is perpendicular to chord AC. If PBA=30, show that PB is parallel to QC.
1249612_cd9fe9af6bce47d386ed3d9be54d8c49.png

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Solution

Given, O is the center of a circle.
AB=AC
OPAB
OQAC
PBA=30
To prove :- BPQC

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=90o[OQAC&OPAB]

AOC=ABO [using (3)]

OC=OB [Radii]

OXCOYB [AAS]

QOC=POB [CPCT] __ (4)

Now in QOC and POB,

OQ=OB[ radius]

QOC=POB (using (4))

OC=OP( radius)

QOCPOB [SAS]

OQC=OBP [CPCT]

These are alternate interior angles formed between QC and PB.
Hence, QCPB[Hence proved]

1192133_1249612_ans_6d6e56b6000348ed8606c43ac94f5049.png

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