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

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Solution

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

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Q. ABCD is a parallelogram, P is a point on side BC such that DP produced meets AB produced at L.. Prove that

(i)

\frac{D P}{P L} = \frac{D C}{B L}

(ii)

\frac{D L}{D P} = \frac{A L}{D C}

Q. In the adjoining figure

, A B C D

is a parallelogram. Any line through A cuts

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Prove that :

a r (Δ B P C) = a r (Δ D P Q) .

1293878_f8df6de59fdb466294cd8e3124285b76.1293878-Q

$P$ is a point on side $B C$ of a parallelogram $A B C D$ . If $D P$ produced meets $A B$ produced at point $L$ , prove that :
(i) $D P : P L = D C : B L$ .
(ii) $D L : D P = A L : D C$ .

In the adjoining figure ABCD is a parallelogram. ‘P’ is a point on BC, DP and AB are produced meet at L. Prove that DP : PL = DC : BL.

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D C

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B C

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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