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Question
In the given figure, AB = AD = DC = PB and ∠DBC = x°. Determine, in terms of x :
- ∠ABD,
- ∠APB.
Hence or otherwise, prove that AP is parallel to DB.
Solution
Given – In the figure, AB = AD = DC = PB and DBC = X°
Join AC and BD
To find : The measure of ∠ABD and ∠APB
Proof : ∠DAC = ∠DBC = X ...[Angels in the same segment]
But ∠DCA = ∠DAC = X ...[∵ AD = DC]
Also, we have, ∠ABD = ∠DAC ...[Angles in the same segment]
In ∆ABP, ext ∠ABC = ∠BAP + ∠APB
But, ∠BAP = ∠APB ...[∵ AB = BP]
2 × X = ∠APB + ∠APB = 2∠APB
∴ 2∠APB = 2X
`=>` ∠APB = X
∴ ∠APB = ∠DBC = X,
But these are corresponding angles
∴ AP || DB
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