ABCD is a a parallelogram A circle passing through D , A,B cuts BC in P Prove that DC - Maths -

Question

ABCD is a a parallelogram A circle passing through D , A,B cuts BC
in P
Prove that DC = DP

Global Expert · Accepted Answer

∠DAB=90° and ∠DPB=90°      Angle in the semicircle is 90°∠DAB+ ∠DPB=180°  Therefore, ABPD is a cyclic quadrilateral
Now, suppose that ∠DAB=θ ∠DAB+∠DPB=180°   (ABPD is a cyclic quadrialteral)⇒∠DPB=180°-θ∠DAB=∠DCB=θ      (Opposite angles of a parallelogram are equal)∠DPC=180°-∠DPB  (Linnear pair)⇒∠DPC=180°-180°+θ⇒∠DPC=θHence, ∠DPC=∠DCP=θi
e
, DC=DP  (Sides opposite to equal angle)

ABCD is a a parallelogram. A circle passing through D , A,B cuts BC in P Prove that DC = DP

ABCD is a a parallelogram. A circle passing through D , A,B cuts BC in P
Prove that DC = DP