# In Fig. 9.27, ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD) [Hint: Join PD].

**Solution:**

Given, ABCD and AEFD are two parallelograms

We have to prove that ar(PEA) = ar(QFD)

Considering triangles APE and DQF,

We know that the corresponding angles are equal

∠APE = ∠DQF

We know that the opposite sides of a parallelogram are equal

AE = DF

∠AEP = ∠DFQ

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 APE and DQF are congruent.

The Corresponding Parts of Congruent Triangles are Congruent (CPCTC) theorem states that when two triangles are comgruent, then their corresponding sides and angles are also congruent or equal in measurements.

By CPCTC,

PE = FQ

We know that congruent triangles have equal area

So, ar(APE) = ar(DQF)

Therefore, ar (PEA) = ar (QFD)

**✦ Try This: **BCD is a parallelogram, G is the point on AB such that AG = 2 GB, E is a point of DC such that CE = 2 DE and F is the point of BC such that BF = 2FC. Prove that: ar(△EGB) = 1/6 ar(ABCD)

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

**NCERT Exemplar Class 9 Maths Exercise 9.4 Problem 10**

## In Fig. 9.27, ABCD and AEFD are two parallelograms. Prove that ar (PEA) = ar (QFD) [Hint: Join PD].

**Summary:**

The opposite sides of a parallelogram are equal in length, and the opposite angles are equal in measure. In Fig. 9.27, ABCD and AEFD are two parallelograms. It is proven that ar (PEA) = ar (QFD)

**☛ Related Questions:**

- In Fig. 9.25, ABCDE is any pentagon. BP drawn parallel to AC meets DC produced at P and EQ drawn par . . . .
- If the medians of a ∆ ABC intersect at G, show that ar(AGB) = ar(AGC) = ar(BGC) = 1/3 ar(ABC)
- In Fig. 9.26, X and Y are the mid-points of AC and AB respectively, QP || BC and CYQ and BXP are str . . . .

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