Question

Plz fast

very urgent

In the figure, two chords AB and CD of the same circle are parallel to each other. P is the centre of the circle Show that angle CPA is congruent with angle DPB

Answer 1

i) In ΔPAC and ΔPDB,

∠P = ∠P (Common)

∠PAC = ∠PDB (Exterior angle of a cyclic quadrilateral is ∠PCA = ∠PBD equal to the opposite interior angle)

∴ ΔPAC ∼ ΔPDB

(ii)We know that the corresponding sides of similar triangles are proportional.

PA/PD=AC/DB =PC/PB
PA/PD =PC/PB

∴ PA.PB = PC.PD

p a y p d is equal to

Answer 2

I am assuming that you have asked that Triangle CPA is similar with Triangle DPB, because angles can't be congruent ☺

ANSWER--

In ΔPAC and ΔPDB,

∠P = ∠P (Common)

∠PAC = ∠PDB (Exterior angle of a cyclic quadrilateral is ∠PCA = ∠PBD equal to the opposite interior angle)

∴ ΔPAC ∼ ΔPDB (By AA similarity criteria)