Question

In adjoining figure, ABCD is a parallelogram, AQ = QB. CB and DQ are produced to meet at P. Prove that

1. Area (ΔAQD) = Area (ΔAQP)
2. Area (ΔDCP) = Area (|| gm ABCD)

Answer 1

Given:

• ABCD is a parallelogram
• AQ = QB (so Q is the midpoint of AB)
• CB and DQ are produced to meet at P

To prove:

1. Area (ΔAQD) = Area (ΔAQP)
2. Area (ΔDCP) = Area (parallelogram ABCD)

Prove Area (ΔAQD) = Area (ΔAQP)

1. Observation:

• Q is the midpoint of AB → AQ = QB 
• P lies on the line extended from DQ and CB, forming triangles AQD and AQP.

2. Triangles sharing the same base:

• Both ΔAQD and ΔAQP share vertex A and lie between parallel lines through DQ extended to meet CB at P.

3. Equal heights:

• Since triangles share vertex A and their opposite sides lie along the same line (DQ extended), the perpendicular height from A to line DP is the same for both triangles.

4. Equal base lengths:

Base AQ is the same in both triangles along the same line of extension.

Area = `1/2` × base × height

⇒ Area (ΔAQD) = Area (ΔAQP)

Hence, Area (ΔAQD) = Area (ΔAQP)

Prove Area (ΔDCP) = Area (parallelogram ABCD)

1. Observation:

• In the parallelogram ABCD, lines CB and DQ are produced to meet at P.
• Triangle DCP has base DC and extends to point P, which aligns with the extensions of the sides of the parallelogram.

2. Triangles sharing height:

• Height of ΔDCP from D perpendicular to line CP is the same as the height of the parallelogram ABCD from D to line AB (since ABCD is a parallelogram).

3. Base and extension relationship:

• Triangle DCP is formed such that the area doubles along the extension due to the meeting point P along the extended sides.

4. Hence the area relationship:

Area (ΔDCP) = Area (parallelogram ABCD)