Question

Show that <CPA=<DPB

Please refer to the image above for full question​

Answer 1

$\Huge{\boxed{\pink{\mathbb{QUESTION}}}}$

In the above figure, two chords AB and CD are || to each other. P is the centre of the circle.

Show that,

∠CPA = ∠DPB

$\Huge{\boxed{\purple{\mathcal{ANSWER}}}}$

Construction =>

Join C and P, A and P, D and P and B and P, C and A, D and B.

In △CPA and △DPB,

CP = DP (Radii of the same circle)

AP = BP (Radii of the same circle)

CA = BD (AB || CD, so distance between them will be equal)

Therefore, △CPA ≅ △DPB

So, ∠CPA = ∠DPB (By C.P.C.T.)

Therefore, hence showed.

Alternate method : -

Construction : - Join C and P, A and P, D and P and B and P.

CP = DP (Radii of the same circle)

AP = BP (Radii of the same circle)

As we know,

Equal chords of a circle substend equal angles at their centre.

Therefore, ∠CPA = ∠DPB