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
Answers
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
