Question

Quadrilateral ABCD is a parallelogram a circle passing through D,A,B cuts BC in p prove that DC is equals to DP​

Answer 1

Answer:

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Step-by-step explanation:

∠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=θ

Hence, ∠DPC=∠DCP=θi.e., DC=DP (Sides opposite to equal angle)

Answer 2

Answer:

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Step-by-step explanation:

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