In the given figure, AB = AD = DC = PB and ∠DBC = x°. Determine, in terms of x :

(i) ∠ABD, (ii) ∠APB.

Hence or otherwise, prove that AP is parallel to DB.

(i) Join AC and BD.

∠DAC = ∠DBC = x° [Angles in the same segment are equal]

∠DCA = ∠DAC = x° [As angles opposite to equal sides are equal]

Also, we have

∠ABD = ∠DCA = x° [Angles in the same segment are equal]

Hence, ∠ABD = x°.

(ii) In ∆ABP

⇒ Ext. ∠ABC = ∠BAP + ∠APB

But, ∠BAP = ∠APB [As angles opposite to equal sides are equal]

⇒ 2x° = ∠APB + ∠APB = 2∠APB

⇒ 2∠APB = 2x°

⇒ ∠APB = x°

Hence, ∠APB = x°.

Thus, ∠APB = ∠DBC = x° [These are corresponding angles]

Hence, proved that AP || DB.