# The diagonals of a parallelogram ABCD intersect at point O. Through O, a line is drawn to intersectAD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal areas.

 Here is the answer to your question. Given: Parallelogram ABCD where the diagonals intersect at O. PQ is a line segment passing through O. To prove: PQ divides the parallelogram ABCD into two parts of equal area. Proof:In ΔBOQ and ΔDOP,OB = OD (Diagonals of parallelogram bisect each other)∠BOQ = ∠DOP (Vertically opposite angles)∠OBQ = ∠ODP (Alternate interior angles)∴ ΔBOQ ≅ ΔDOP (ASA Congruence criterion)⇒ Area (ΔBOQ) = Area (ΔDOP) … (1)The diagonal BD of parallelogram ABCD divides it into two triangles ΔBCD and ΔABD of equal areasArea (ΔBCD) = Area (ΔABD) … (2) Now, area (QCDP) = area (ΔBCD) + area (ΔDOP) - area (ΔBOQ)⇒ area (QCDP) = area (ΔABD) + area (ΔBOQ) - area (ΔDOP) [Using (1) and (2)]= area (APQB):) :)
• 90

 Here is the answer to your question. Given: Parallelogram ABCD where the diagonals intersect at O. PQ is a line segment passing through O. To prove: PQ divides the parallelogram ABCD into two parts of equal area. Proof:In ΔBOQ and ΔDOP,OB = OD (Diagonals of parallelogram bisect each other)∠BOQ = ∠DOP (Vertically opposite angles)∠OBQ = ∠ODP (Alternate interior angles)∴ ΔBOQ ≅ ΔDOP (ASA Congruence criterion)⇒ Area (ΔBOQ) = Area (ΔDOP) … (1)The diagonal BD of parallelogram ABCD divides it into two triangles ΔBCD and ΔABD of equal areasArea (ΔBCD) = Area (ΔABD) … (2) Now, area (QCDP) = area (ΔBCD) + area (ΔDOP) - area (ΔBOQ)⇒ area (QCDP) = area (ΔABD) + area (ΔBOQ) - area (ΔDOP) [Using (1) and (2)]= area (APQB)
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Given: Parallelogram ABCD where the diagonals intersect at O. PQ is a line segment passing through O.

To prove: PQ divides the parallelogram ABCD into two parts of equal area.

Proof:

In ΔBOQ and ΔDOP,

OB = OD (Diagonals of parallelogram bisect each other)

∠BOQ = ∠DOP (Vertically opposite angles)

∠OBQ = ∠ODP (Alternate interior angles)

∴ ΔBOQ ≅ ΔDOP (ASA Congruence criterion)

⇒ Area (ΔBOQ) = Area (ΔDOP) … (1)

The diagonal BD of parallelogram ABCD divides it into two triangles ΔBCD and ΔABD of equal areas

Area (ΔBCD) = Area (ΔABD) … (2)

Now, area (QCDP) = area (ΔBCD) + area (ΔDOP) - area (ΔBOQ)

⇒ area (QCDP) = area (ΔABD) + area (ΔBOQ) - area (ΔDOP) [Using (1) and (2)]

= area (APQB)
• 13 • 3 • -6
Here is the answer to your question. Given: Parallelogram ABCD where the diagonals intersect at O. PQ is a line segment passing through O. To prove: PQ divides the parallelogram ABCD into two parts of equal area. Proof: In ΔBOQ and ΔDOP, OB = OD (Diagonals of parallelogram bisect each other) ∠BOQ = ∠DOP (Vertically opposite angles) ∠OBQ = ∠ODP (Alternate interior angles) ∴ ΔBOQ ≅ ΔDOP (ASA Congruence criterion) ⇒ Area (ΔBOQ) = Area (ΔDOP) … (1) The diagonal BD of parallelogram ABCD divides it into two triangles ΔBCD and ΔABD of equal areas Area (ΔBCD) = Area (ΔABD) … (2) Now, area (QCDP) = area (ΔBCD) + area (ΔDOP) - area (ΔBOQ) ⇒ area (QCDP) = area (ΔABD) + area (ΔBOQ) - area (ΔDOP) [Using (1) and (2)] = area (APQB) That the answer .....😃😃😃😃
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Whole root 30 + whole root 30 + root 30+ so on evaluate
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Diagnols of paeallgram pqrs • 0
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