Answer :

i) Construction: Join P to B to Q

Join AB, we get,

In, blue circle, ∠ABP = 90° (since, angle in semicircle is right angle)

In, green circle, ∠ABQ = 90° (since, angle in semicircle is right angle)

Thus, ∠ABP + ∠ABQ = 90 + 90 =180°

Hence, P, B, Q lie on a line.

ii) In the above diagram,

iv. Join, center O and O’ and OB and O’B, we get

Let D be the point of intersection of OO’ and AB

In Δ AOO’ and Δ BOO’

AO = BO

(radius of a blue circle )

AO’ = BO’

( radius of a green circle )

OO’ = OO’

(common side)

∴ Δ AOO’ ≅ Δ BOO’

(by SSS congruent rule )

∠AOO’ = ∠BOO’ (by CPCT) ….(1)

In Δ AOD and Δ BOD

AO = BO (radius of a blue circle )

∠AOO’ = ∠BOO’ ( from (1) )

OD = OD (common side)

∴ Δ AOD ≅ Δ BOD (by SAS congruent rule )

∠ODA = ∠ODB (by CPCT) ….(2)

Since, sum of angles on a line is 180°

∴ ∠ODA + ∠ODB = 180 (∵ AB is a line)

⇒ ∠ODA + ∠ODA = 180 (from (2))

⇒ 2∠ODA = 180

…..(3)

⇒ ∠ODB = ∠ODA = 90° ( from (2) and (3))

As AB is a transversal line,

Also, We know that, In, blue circle, ∠ABP = 90° (since, angle in

semicircle is right angle)

And ∠ODB = 90°

∵ ∠ABP and ∠ODB are co interior angles.

And, as ∠ABP + ∠ODB = 90 + 90 =180°

We know that, if sum of cointerior angle is 180° then the line is parallel.

Therefore, OO’ is parallel to PQ.

⇒ OO’ ∥ PQ …..(4)

To show: PQ = 2 OO’

In Δ APQ, O and O’ is the mid point of line AP and AQ respectively (since, O and O’ are centre of a circle)

Also, from (4)

OO’ ∥ PQ

Thus, by ** Mid point theorem:** The line which joins the midpoints of two sides of a triangle is parallel to the third side and is equal to half of the length of the third side.

We get,

⇒ PQ = 2 OO’

Hence proved.

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