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))


2ODA = 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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