In the given figure, AB = AD = DC = PB and ∠DBC = xo. Determine, in terms of x: (i) ∠ABD, (ii) ∠APB. Hence or otherwise, prove that AP is parallel to DB. - Noon Academy

In the given figure, AB = AD = DC = PB and ∠DBC = xo. Determine, in terms of x: (i) ∠ABD, (ii) ∠APB. Hence or otherwise, prove that AP is parallel to DB.

Circles | Class 10 | Exercise 17C | ICSE | Maths | Selina

In the given figure, AB = AD = DC = PB and ∠DBC = xo. Determine, in terms of x: (i) ∠ABD, (ii) ∠APB. Hence or otherwise, prove that AP is parallel to DB.

Selina Solutions Concise Class 10 Maths Chapter 17 ex. 17(C) - 13

Solution:

Selina Solutions Concise Class 10 Maths Chapter 17 ex. 17(C) - 14

Given,

$AB\text{ }=\text{ }AD\text{ }=\text{ }DC\text{ }=\text{ }PB\text{ }and~\angle DBC\text{ }=\text{ }{{x}^{o}}$

Join

$AC\text{ }and\text{ }BD$

Proof:

$\angle DAC\text{ }=\angle DBC\text{ }=\text{ }{{x}^{o}}$

[Angles in the same segment]

And,

$\angle DCA\text{ }=\angle DAC\text{ }=\text{ }{{x}^{o}}~$

$\left[ As\text{ }AD\text{ }=\text{ }DC \right]$

Also, we have

$\angle ABD\text{ }=\angle DAC$

[Angles in the same segment]

And, in

$\vartriangle ABP$

$Ext.\angle ABD\text{ }=\angle BAP\text{ }+\angle APB$

But,

$\angle BAP\text{ }=\angle APB$

$\left[ Since,\text{ }AB\text{ }=\text{ }BP \right]$

$2\text{ }{{x}^{o}}~=\angle APB\text{ }+\angle APB\text{ }=\text{ }2\angle APB$

$2\angle APB\text{ }=\text{ }2{{x}^{o}}$

So,

$\angle APB\text{ }=\text{ }{{x}^{o}}$

Thus,

$\angle APB\text{ }=\angle DBC\text{ }=\text{ }{{x}^{o}}$

But these are corresponding angles,

Therefore,

$AP\text{ }||\text{ }DB$

i

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