Question

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Answer 1

Answer:

Refer to the image

Step-by-step explanation:

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Answer 2

Step-by-step explanation:

$\mathsf{Given:- DP = BQ}$

$\mathsf{Proof:-}$

$\mathsf{(i) }$

$\mathsf{AD=BC (Opposite \: side \: of \: parallelogram)}$

$\mathsf{DP=BQ (Given)}$

$\mathsf{∠ ADP=∠QBC ( Alternate \: angles \: parallel \: sides)}$

$\mathsf{By \: SAS}$

$\mathsf{∴∆APD≅∆CQB}$

$\mathsf{(ii)}$

$\mathsf{By \: CPCT - Corresponding \: parts \: of \: congruent \: triangle}$

$\mathsf{∴AP=CQ}$

$\mathsf{(iii) }$

$\mathsf{AB=CD (Opposite \: sides \: of \: parallelogram)}$

$\mathsf{DP=BQ (Given)}$

$\mathsf{∠PDC=ABQ ( Alternate \: angles \: of \: parallel \: sides)}$

$\mathsf{By \: SAS}$

$\mathsf{∴∆AQB≅∆ACPD}$

$\mathsf{(iv)}$

$\mathsf{By \: CPCT- Corresponding \: parts \: of \: congruent \: triangle}$

$\mathsf{AQ=CP}$

$\mathsf{AP=CQ \: \: \: This \: are \: equal.}$

$\mathsf{AQ=CP \: This \: are \: equal}$

$\mathrm{∴APCQ \: is \: a \: parallelogram.}$