Question

if two circles touch internally at the point p and a line ABCD intersection the outer circle at the point A and D and intersection the inner circle at the point B and C. prove that angle APB= angle CPD

Answer 1

angle APB is proved to be equal to angle CPD if two circles touch internally at P and a straight line ABCD meets the outer circle in a and D and the inner circle in B and C.

Step-by-step explanation:

It is given that, (Referring to the figure attached below)

Two circles touch internally at P

ABCD is a straight line that meets the outer circle at A & D and the inner circle at B & C.

Let’s draw a line XPY in such a way that it forms a common tangent to both the circles at P.

Let’s assume from the figure,

∠XPA = ∠2

∠APB = ∠1

∠PCB = ∠3

∠CPD = ∠4

∠PDC = ∠5

We know that according to the alternate segment theorem the angle between a chord and a tangent through one of the endpoints of the chord is equal to the angle in the alternate segment.

So, for the chord AP in the outer circle and the tangent XY, we have

∠XPA = ∠PDC

⇒ ∠2 = ∠5 …. (i)

Similarly, for the chord BP in the inner circle and the tangent XY, we have

∠XPB = ∠PCB  

⇒ ∠XPA + ∠APB = ∠PCB

⇒ ∠2 +  ∠1 = ∠3 …… (ii)

Considering ∆ CPD, we have  

∠CPD + ∠PDC = ∠PCB ….. [exterior angle is equal to the sum of two opposite interior angles]

⇒ ∠4 + ∠5 = ∠3 …… (iii)

Therefore, from (ii) & (iii), we get

∠2 + ∠1 = ∠4 + ∠5

⇒ ∠1 = ∠4 …… [since in eq. (i) ∠2 = ∠5]

⇒ ∠APB = ∠CPD

Hence proved

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