In Fig. 9.32, 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.]

Solution:

It is given that ABCD is a parallelogram.

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

Now, join the points A and C.

Consider ΔAPC and ΔBPC

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

According to Theorem 9.2: Two triangles on the same base (or equal bases) and between the same parallels are equal in area.

Therefore,

Area (ΔAPC) = Area (ΔBPC) ...(1)

In quadrilateral ACQD, 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 that is obtained when line segment BC is produced.

∴ AD || CQ

We have,

AD = CQ and AD|| CQ

Hence, ACQD is a parallelogram.

Consider ∆DCQ and ∆ACQ

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) [Subtracting Area (ΔPQC) on both sides.]

∴ Area (ΔDPQ) = Area (ΔAPC) ...(2)

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

Area (ΔBPC) = Area (ΔDPQ), proved.

☛ Check: NCERT Solutions for Class 9 Maths Chapter 9

---

Video Solution:

In Fig. 9.32, 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.]

Maths NCERT Solutions Class 9 Chapter 9 Exercise 9.4 Question 4

Summary:

If ABCD is parallelogram and BC is produced to a point Q such that AD = CQ and AQ intersect DC at P in the given figure, then ar (ΔBPC) = ar (ΔDPQ).

---

☛ Related Questions:

• In Fig.9.33, ABC and BDE are two equilateral triangles such that D is the mid-point of BC. If AE intersects BC at F, show thati) ar (BDE) =1/4 ar (ABC)ii) ar (BDE) = 1/2 ar (BAE)iii) ar (ABC) = 2 ar (BEC)iv) ar (BFE) = ar (AFD)v) ar (BFE) = 2 ar (FED)vi) ar (FED) = 1/8 ar (AFC)[Hint : Join EC and AD. Show that BE || AC and DE || AB, etc.]
• Diagonals AC and BD of a quadrilateral ABCD intersect each other at P. Show that ar (ΔAPB) × ar (ΔCPD) = ar (ΔAPD) × ar (ΔBPC)[Hint: From A and C, draw perpendiculars to BD.]
• P and Q are respectively the mid-points of sides AB and BC of a triangle ABC and R is the mid-point of AP, show thati) ar (PRQ) = 1/2 ar (ARC)ii) ar (RQC) = 3/8 ar (ABC)iii) ar (PBQ) = ar (ARC)
• In Fig. 9.34, ABC is a right triangle right angled at A. BCED, ACFG and ABMN are squares on the sides BC, CA and AB respectively. Line segment AX ⊥ DE meets BC at Y. Show that:i) DMBC ≅ DABDii) ar (BYXD) = 2ar (MBC)iii) ar (BYXD) = ar (ABMN)iv) ΔFCB ≅ ΔACEv) ar (CYXE) = 2ar (FCB)vi) ar (CYXE) = ar (ACFG)vii) ar (BCED) = ar (ABMN) + ar (ACFG)Note: Result (vii) is the famous Theorem of Pythagoras. You shall learn a simpler proof of this theorem in class X.