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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Answered by
8
don't worry this question is simple.
just show the chords are parallel and the line joining the centre of the circle and the chord is their transversal. hence the angle which we have to find are alternate angles.
hence by alternate angle property,
the angles are congruent.
just write this in mathematical form. you will get full marks by solving by this method
just show the chords are parallel and the line joining the centre of the circle and the chord is their transversal. hence the angle which we have to find are alternate angles.
hence by alternate angle property,
the angles are congruent.
just write this in mathematical form. you will get full marks by solving by this method
Answered by
5
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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