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In the figure, two chords AB and CD of the same circle are parallel to each other. P is the centre of the circle Show that angle CPA is congruent with angle DPB
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(i) In ΔPAC and ΔPDB,
∠P = ∠P (Common)
∠PAC = ∠PDB (Exterior angle of a cyclic quadrilateral is ∠PCA = ∠PBD equal to the opposite interior angle)
∴ ΔPAC ∼ ΔPDB
∠P = ∠P (Common)
∠PAC = ∠PDB (Exterior angle of a cyclic quadrilateral is ∠PCA = ∠PBD equal to the opposite interior angle)
∴ ΔPAC ∼ ΔPDB
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QUESTION :
P is the centre of the circle . Two chords AB and CD are parallel to each other.To prove that angle CPA is congruent to angle DPB
ANSWER :
GIVEN :
In a circle at centre P
AB || CD
TO PROVE :
angle CPA = angle DPB
OR
angle CPA is congruent to angle DPB
CONSTRUCTION :
join AP,CP,DP,BP
and draw PE perpendicular to AB
and PF perpendicular to CD
such that EF is a line
PROOF :
now in ∆AEP and ∆DPF
angle AEP = angle DFP.......(each 90°)
also angle APE = angleDPF .....(vertically opposite angles)
also PA = PD.......(radius of same circle)
so by AA congruence criteria
∆AEP is congruent to ∆DPF
so by CPCT
angle PAE = angle PDF.....1
AND
now in ∆CFP and ∆BEP
angle CFP = angle BEP.......(each 90°)
also angle CPF = angleBPE .....(vertically opposite angles)
also PC = PB......(radius of same circle)
so by AA congruence criteria
∆CPF is congruent to ∆BPE
so by CPCT
angle PCF = angle PBE......2
NOW
consider ∆ CDF
by exterior angle theorem
angle CPA = angle PCD + angle PDC.....3
Similarly in ∆APB
by exterior angle theorem
angle BPD = angle PAB + angle PBA......4
but from 1 and 2
angle PAE = angle PDF = angle PDC
and angle PCF = angle PBE = angle PBA
so equation 4 becomes
angle BPD = angle PAB + angle PBA
angle BPD = angle PDC + angle PCD....5
but from 3
angle CPA = angle PCD + angle PDC
SO equation 5 becomes
angle BPD = angle PDC + angle PCD
angle BPD = angle CPA
HENCE PROVED
NOTE :
while solving such questions
try to construct the needed things
P is the centre of the circle . Two chords AB and CD are parallel to each other.To prove that angle CPA is congruent to angle DPB
ANSWER :
GIVEN :
In a circle at centre P
AB || CD
TO PROVE :
angle CPA = angle DPB
OR
angle CPA is congruent to angle DPB
CONSTRUCTION :
join AP,CP,DP,BP
and draw PE perpendicular to AB
and PF perpendicular to CD
such that EF is a line
PROOF :
now in ∆AEP and ∆DPF
angle AEP = angle DFP.......(each 90°)
also angle APE = angleDPF .....(vertically opposite angles)
also PA = PD.......(radius of same circle)
so by AA congruence criteria
∆AEP is congruent to ∆DPF
so by CPCT
angle PAE = angle PDF.....1
AND
now in ∆CFP and ∆BEP
angle CFP = angle BEP.......(each 90°)
also angle CPF = angleBPE .....(vertically opposite angles)
also PC = PB......(radius of same circle)
so by AA congruence criteria
∆CPF is congruent to ∆BPE
so by CPCT
angle PCF = angle PBE......2
NOW
consider ∆ CDF
by exterior angle theorem
angle CPA = angle PCD + angle PDC.....3
Similarly in ∆APB
by exterior angle theorem
angle BPD = angle PAB + angle PBA......4
but from 1 and 2
angle PAE = angle PDF = angle PDC
and angle PCF = angle PBE = angle PBA
so equation 4 becomes
angle BPD = angle PAB + angle PBA
angle BPD = angle PDC + angle PCD....5
but from 3
angle CPA = angle PCD + angle PDC
SO equation 5 becomes
angle BPD = angle PDC + angle PCD
angle BPD = angle CPA
HENCE PROVED
NOTE :
while solving such questions
try to construct the needed things
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