In Figure, ABCD is a parallelogram and BC is produced to a point Q such that AD = CQ. If AQ intersect DC at P, show that ar (BPC) = ar (DPQ). [Hint : Join AC.]

(i) Let G and H be the mid-points of side AB and AC respectively.

Line segment GH is joining the mid-points. Therefore, it will be parallel to third side BC and also its length will be half of the length of BC (mid-point theorem).

⇒ GH = BC and GH || BD

⇒ GH = BD = DC and GH || BD (D is the mid-point of BC)

Consider quadrilateral GHDB.

GH ||BD and GH = BD

Two line segments joining two parallel line segments of equal length will also be equal and parallel to each other.

Therefore, BG = DH and BG || DH

Hence, quadrilateral GHDB is a parallelogram.

We know that in a parallelogram, the diagonal bisects it into two triangles of equal area.

Hence, Area (ΔBDG) = Area (ΔHGD)

Similarly, it can be proved that quadrilaterals DCHG, GDHA, and BEDG are parallelograms and their respective diagonals are dividing them into two triangles of equal area.

ar (ΔGDH) = ar (ΔCHD) (For parallelogram DCHG)

ar (ΔGDH) = ar (ΔHAG) (For parallelogram GDHA)

ar (ΔBDE) = ar (ΔDBG) (For parallelogram BEDG)

ar (ΔABC) = ar(ΔBDG) + ar(ΔGDH) + ar(ΔDCH) + ar(ΔAGH)

ar (ΔABC) = 4 × ar(ΔBDE)

Hence, 

(ii)Area (ΔBDE) = Area (ΔAED) (Common base DE and DE||AB)

Area (ΔBDE) − Area (ΔFED) = Area (ΔAED) − Area (ΔFED)

Area (ΔBEF) = Area (ΔAFD) (1)

Area (ΔABD) = Area (ΔABF) + Area (ΔAFD)

Area (ΔABD) = Area (ΔABF) + Area (ΔBEF) [From equation (1)]

Area (ΔABD) = Area (ΔABE) (2)

AD is the median in ΔABC.

From (2) and (3), we obtain

2 ar (ΔBDE) = ar (ΔABE)

Or, 

(iii) 

ar (ΔABE) = ar (ΔBEC) (Common base BE and BE||AC)

ar (ΔABF) + ar (ΔBEF) = ar (ΔBEC)

Using equation (1), we obtain

ar (ΔABF) + ar (ΔAFD) = ar (ΔBEC)

ar (ΔABD) = ar (ΔBEC)

ar (ΔABC) = 2 ar (ΔBEC)

(iv)It is seen that ΔBDE and ar ΔAED lie on the same base (DE) and between the parallels DE and AB.

∴ar (ΔBDE) = ar (ΔAED)

⇒ ar (ΔBDE) − ar (ΔFED) = ar (ΔAED) − ar (ΔFED)

∴ar (ΔBFE) = ar (ΔAFD)

(v)Let h be the height of vertex E, corresponding to the side BD in ΔBDE.

Let H be the height of vertex A, corresponding to the side BC in ΔABC.

In (i), it was shown that

In (iv), it was shown that ar (ΔBFE) = ar (ΔAFD).

∴ ar (ΔBFE) = ar (ΔAFD)

= 

= 2 ar (ΔFED)

Hence, 

(vi) Area (AFC) = area (AFD) + area (ADC)

Now, by (v), … (6)

Therefore, from equations (5), (6), and (7), we get:

It is given that ABCD is a parallelogram.

AD || BC and AB || DC(Opposite sides of a parallelogram are parallel to each other)

Join point A to point C.

Consider ΔAPC and ΔBPC

ΔAPC and ΔBPC are lying on the same base PC and between the same parallels PC and AB. Therefore,

Area (ΔAPC) = Area (ΔBPC) ... (1)

In quadrilateral ACDQ, it is given that

AD = CQ

Since ABCD is a parallelogram,

AD || BC (Opposite sides of a parallelogram are parallel)

CQ is a line segment which is obtained when line segment BC is produced.

∴ AD || CQ

We have,

AC = DQ and AC || DQ

Hence, ACQD is a parallelogram.

Consider ΔDCQ and ΔACQ

These are on the same base CQ and between the same parallels CQ and AD. Therefore,

Area (ΔDCQ) = Area (ΔACQ)

⇒ Area (ΔDCQ) − Area (ΔPQC) = Area (ΔACQ) − Area (ΔPQC)

⇒ Area (ΔDPQ) = Area (ΔAPC) ... (2)

From equations (1) and (2), we obtain

Area (ΔBPC) = Area (ΔDPQ)