Question

if two circles are internally touching at point P and a line intersect those two line in points A,B,C,D respectively then prove that angle APB is equals to angle CPD

Answer 1

Proved that ∠APB = ∠CPD.    

Given

To prove that, ∠APB = ∠CPD

From the figure,

In a straight line ABCD, A and D cuts at outer circle  and

                                       B and C cuts at inner circle.

Two circles touches internally at point P.

Now, draw a line XPY, which is the common tangent to the circles at P.

Hence, the angles are in alternate segment are equal.

             ∠APB = ∠1  

             ∠XPA = ∠2

             ∠PCB = ∠3

             ∠CPD = ∠4

             ∠PDC = ∠5

Outer circle for chord AP,

             ∠XPA = ∠PDC

             ∠2  =  ∠5    -----> ( 1 )

Inner circle for chord BP,

             ∠XPB = ∠PCB

                 ∠1 + ∠2 = ∠3   -----> ( 2 )

In ΔCPD,

Exterior angle is equal to the sum of the opposite interior angle,

              ∠CPD + ∠PDC = ∠PCB

              ∠4 + ∠5  =  ∠3   -----> ( 3 )

From the equation ( 2 ) and ( 3 ),

               ∠1 + ∠2 = ∠4 + ∠5

               ∠1 + ∠5 = ∠4 + ∠5

                       ∠1 = ∠4

Therefore, it has been proved that ∠APB = ∠CPD.

To learn more...

1. brainly.in/question/2447170

2. brainly.in/question/2021577