Question

two circle with Centre P and Q touch externally at point A such in that PAQ is a straight line theough A a line is drawn to intersect at A B and C are parallel

Answer 1

kindly see attachment for fig.

Given: Two circles with centre A and B touches at P and CD passing through P

 

In ΔACP

AC = AP (radius) 

⇒ ∠APC = ∠ACP  (Angles opposite to equal sides )  ...  (1)

 

In ΔBDP

BD = BP (radius) 

⇒ ∠BPD = ∠BDP  ...  (2)

 

But

∠APC = ∠BPD  (Vertically opposite angles)  ...  (3)

Now AC and DB are two lines and AB is the transversal such that 

∠ACP = ∠BPD          ( from (1), (2) and (3) )

 

Hence AC || BD

Answer 2

Answer:

Step-by-step explanation:

Given: Two circles with centre A and B touches at P and CD passing through P

 

In ΔACP

AC = AP (radius)  

⇒ ∠APC = ∠ACP  (Angles opposite to equal sides )  ...  (1)

 

In ΔBDP

BD = BP (radius)  

⇒ ∠BPD = ∠BDP  ...  (2)

 

But

∠APC = ∠BPD  (Vertically opposite angles)  ...  (3)

Now AC and DB are two lines and AB is the transversal such that  

∠ACP = ∠BPD          ( from (1), (2) and (3) )

 

Hence AC || BD