plz solve question 11
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see the diagram.
Let O be the center of the circle. Join OB and OD.
Draw OE ⊥AB. and OF⊥ CD.
Equal chords are at equal distances from center O. Hence OE = OF.
EB = AB/2 = CD/2 = FD.
OB = OD = radius.
Hence ΔOEB and ΔOFD are congruent.
So ∠OBA = ∠ODC.
In the ΔOBD, OB = OD, so ∠OBD = ∠ODB .
So ∠PBD = 180 - ∠OBD - ∠OBA
= 180° - ∠ODB - ∠ODC
= ∠PDB
Hence, in the Isosceles ΔPBD, PB = PD.
Let O be the center of the circle. Join OB and OD.
Draw OE ⊥AB. and OF⊥ CD.
Equal chords are at equal distances from center O. Hence OE = OF.
EB = AB/2 = CD/2 = FD.
OB = OD = radius.
Hence ΔOEB and ΔOFD are congruent.
So ∠OBA = ∠ODC.
In the ΔOBD, OB = OD, so ∠OBD = ∠ODB .
So ∠PBD = 180 - ∠OBD - ∠OBA
= 180° - ∠ODB - ∠ODC
= ∠PDB
Hence, in the Isosceles ΔPBD, PB = PD.
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kvnmurty:
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