# In Fig.9.11, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR = RS and PA || QB || RC. Prove that ar (PQE) = ar (CFD).

**Solution:**

Given, PSDA is a parallelogram

P and Q are taken such that PQ = QR = RS

PA || QB || RC

We have to prove that ar(PQE) = ar(CFD)

We know that the opposite sides of a parallelogram are parallel and equal.

PS = AD

PS = PQ + QR + RS

Also, PQ = QR = RS

AD = AB + BC + CD

Also, AB = BC = CD

So, 1/3 PS = 1/3 AD

PQ = CD --------------- (1)

Since PS || AD and cut by a transversal QB,

We know that the alternate interior angles are equal.

∠PQE = ∠CBE ---------------- (2)

Similarly, QB || RC and cut by a transversal AD,

We know that the corresponding angles are equal.

∠QBD = ∠RCD ----------------- (3)

Also, ∠QPE = ∠FCD

Comparing (2) and (3),

∠PQE = ∠QBD

∠RCD = ∠FCD

Considering triangles PQE and CFD,

We know that the alternate angles are equal

∠PQE = ∠CDF

PQ = CD

From (4), ∠QPE = ∠FCD

The Angle-Side-Angle Postulate (ASA) states that if two angles and the included side of one triangle are congruent to two angles and the included side of another triangle, then the two triangles are congruent.

By ASA criteria, the triangles PQE and CFD are congruent.

We know that congruent triangles have equal areas.

Therefore, ar(PQE) = ar(CFD)

**✦ Try This:** In trapezium ABCD ,side AB∣∣DC , diagonals AC and BD intersect in point O. If AB = 20,DC = 6,OB = 15 then Find OD.

**☛ Also Check: **NCERT Solutions for Class 9 Maths Chapter 9

**NCERT Exemplar Class 9 Maths Exercise 9.3 Problem 1**

## In Fig.9.11, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR = RS and PA || QB || RC. Prove that ar (PQE) = ar (CFD).

**Summary:**

In Fig.9.11, PSDA is a parallelogram. Points Q and R are taken on PS such that PQ = QR = RS and PA || QB || RC. It is proven that ar (PQE) = ar (CFD) by ASA criterion

**☛ Related Questions:**

- X and Y are points on the side LN of the triangle LMN such that LX = XY = YN. Through X, a line is d . . . .
- The area of the parallelogram ABCD is 90 cm² (see Fig.9.13). Find (i) ar (ABEF), (ii) ar (ABD), (iii . . . .
- In ∆ ABC, D is the mid-point of AB and P is any point on BC. If CQ || PD meets AB in Q (Fig. 9.14), . . . .

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