O is the centre of the circle in which Seg AB and seg AC are congruent chords. Radius OP is perpendicular to chord AB and radius OQ is perpendicular to chord AC. If angle PBA=30°,show that seg PB is parallel to seg QC. Please solve immediately.
Answers
Solution:
In a circle with center O , AB and AC are two equal chords.Line segment (OM P) ⊥ AB and Line segment (ON Q) ⊥ AC .
Coming to the quadrilateral AMNO
∠OMA=∠ONA→→→Each being 90°
As, AB = CA→→(Given, chords of equal length)
So, AN = AM→→[ ⊥ from the center to the chord bisects the chord]
and, OM= ON→→[ Equal chords are equidistant from the center]
Quadrilateral AMNO is a square.
In ΔPMB
∠PBM + ∠PMB +∠MPB=180°→→Angle sum property of triangle.
30° + 90°+∠MPB=180°
→→∠MPB=60°
In Δ PMB and Δ QNC
→→∠QNC= ∠PMB= 90°
→QN = PM→→→[OQ=OP, also OM=ON→OQ-ON=OP-OM→→QN=PM]
→MB = NC→→[ ⊥ from the center to the chord bisects the chord, also AB=AC, so CN=MB]
→Δ PMB ≅ Δ QNC→→→[SAS]
∠MBP=∠NQC=30°→→→[CPCT]
Join PQ.
OP=OQ= radii of circle
In Δ QOP
∠OQP +∠OPQ +∠POQ=180°
2∠OQP + 90°= 180°
2∠OQP = 180°-90°
2∠OQP = 90°
∠OQP= 45°
Now, ∠QPB +∠CQP=60°+45°+45°+30°=180
Shows that, Segment PB ║ Segment QC →→If sum of interior angles on the same side of transversal is 180°, then lines are parallel]
