# Line AB and CD are parallel and P is any point between them. Show that angleABP + angleCDP = angleDPB

Given AB || CD.

Let EF be the parallel line to AB and CD which passes through P. It can be seem from the figure

∠ABP = ∠BPF  (Alternate int. angles)

∠CDP = ∠DPF  (Alternate int. angles)

⇒ ∠ABP + ∠CDP = ∠BPF + ∠DPF

⇒ ∠ABP + ∠CDP = ∠DPB

Hence proved!!

• 5

for proving it...first u need to construct a parallel line passing through point P between line  AB and CD...so from here you will get

(alternate interior angles between ||lines AB & CD)-
angleABP = angleBPF     ------(i)
angleCDP = angleDPF    -----(ii)

adding (i) & (ii) we get-
angleABP + angleCDP = angleBPF + angleBPD

angleABP + angleCDP = angleBPD              .....-------------hence, proved!!!

i wud sugest u to draw figure first...

• 6

for proving it...first u need to construct a parallel line ,name EF passing through point P between line AB and CD...so from here you will get

(alternate interior angles between ||lines AB & CD)-
angleABP = angleBPF ------(i)
angleCDP = angleDPF -----(ii)

adding (i) & (ii) we get-
angleABP + angleCDP = angleBPF + angleBPD

angleABP + angleCDP = angleBPD .....-------------hence, proved!!!

i wud sugest u to draw figure first...

• -6
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