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 1

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 :

<CPA = <DPB

OR

<CPA is congruent to <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

<AEP = <DFP.......(each 90°)

also <APE = <DPF .....(vertically opposite angles)

also PA = PD.......(radius of same circle)

so by AAS congruence criteria

∆AEP is congruent to ∆DPF

so by CPCT

<PAE = <PDF.....1

AND

now in ∆CFP and ∆BEP

<CFP = <BEP.......(each 90°)

also <CPF = <BPE .....(vertically opposite angles)

also PC = PB......(radius of same circle)

so by AAS congruence criteria

∆CPF is congruent to ∆BPE

so by CPCT

<PCF = <PBE......2

NOW

consider ∆ CDF

by exterior angle theorem

<CPA = <PCD + < PDC.....3

Similarly in ∆APB

by exterior angle theorem

<BPD = <PAB + < PBA......4

but from 1 and 2

<PAE = <PDF = < PDC

and < PCF = < PBE = < PBA

so equation 4 becomes

<BPD = <PAB + <PBA

<BPD = < PDC + < PCD....5

but from 3

<CPA = < PCD + < PDC

SO equation 5 becomes

<BPD = < PDC + < PCD

<BPD = <CPA

HENCE PROVED

NOTE :

while solving such questions

try to construct the needed things

Answer 2

Answer:

Answer is in the attachment