The diagonals of a parallelogram ABCD intersect at point O. Through O, a line is drawn to intersect

AD at P and BC at Q. Show that PQ divides the parallelogram into two parts of equal areas.

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)

Please find this answer

Please find this answer

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 .....😃😃😃😃

Whole root 30 + whole root 30 + root 30+ so on evaluate

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