Question

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very urgent

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

Answer 1

(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

Answer 2

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