NCERT Class 9 Quadrilateral Exercise 8.2 Solutions

Q1

NUMERIC2 marks

ABCD is a quadrilateral in which P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. AC is a diagonal. Show that:

(i) \(\small SR || AC\) and \(SR = \frac{1}{2}AC\)
(ii) \(\small PQ = SR\)
(iii) \(\small PQRS\) is a parallelogram.

📘 Concept & Theory Theory and Concepts Used ›

Midpoint Theorem: In a triangle, the line segment joining the midpoints of two sides is parallel to the third side and equal to half of the third side.
If one pair of opposite sides of a quadrilateral are equal and parallel, then the quadrilateral is a parallelogram.
Midpoint theorem is one of the most important geometric tools used in quadrilaterals, triangles and coordinate geometry.

🗺️ Solution Roadmap Step-by-step Plan ›

Apply midpoint theorem in triangle \(\small ACD\) to establish relation between \(\small SR\) and \(\small AC\).
Apply midpoint theorem in triangle \(\small ABC\) to establish relation between \(\small PQ\) and \(\small AC\).
Compare the two results to prove \(\small PQ = SR\).
Since opposite sides become equal and parallel, conclude that \(\small PQRS\) is a parallelogram.

📊 Graph / Figure Graph / Figure ›

Fig. 8.20

📐 Proof Proof ›

📌 Given

\(\small P,\;Q,\;R,\;S\) are the mid-points of \(\small AB,\;BC,\;CD,\;DA\) respectively and diagonal \(\small AC\) is drawn.

🎯 To Prove

\(\small SR \parallel AC \)

\(\small SR=\frac{1}{2}AC \)

\(\small PQ=SR \)

\(\small PQRS \text{ is a parallelogram} \)

✍️ Proof

Step-by-step Proof · 17 steps · 3 parts

Part (i) - Prove that \(\small SR || AC\) and \(\small SR=\frac{1}{2}AC\)
Consider triangle \(ACD\).
Since \(\small S\) is the midpoint of \(\small AD\),\[\small AS = SD\]
Since \(\small R\) is the midpoint of \(\small CD\),\[\small CR = RD\]
Therefore, points \(\small S\) and \(\small R\) are the midpoints of two sides of triangle \(\small ACD\).
By the Midpoint Theorem,\[\small SR \parallel AC\] and \[\small SR = \frac{1}{2}AC\]
Part - (ii)
Prove that \(\scriptsize PQ = SR\)
Now consider triangle \(\small ABC\).
Since \(\small P\) is the midpoint of \(\small AB\),\[\small AP = PB\]
Since \(\small Q\) is the midpoint of \(\small BC\),\[\small BQ = Q\]
Therefore, \(\small P\) and \(\small Q\) are the midpoints of two sides of triangle \(\small ABC\).
Applying Midpoint Theorem,\[\small PQ \parallel AC\] and \[\small PQ = \frac{1}{2}AC\]
From part (i),\[\small SR = \frac{1}{2}AC\]
Therefore,\[\small PQ = SR\]
Part (iii) - Prove that \(\small PQRS\) is a parallelogram
From parts (i) and (ii),\[\small PQ \parallel AC\] and \[\small SR \parallel AC\]
Since both \(PQ\) and \(SR\) are parallel to \(\small AC\),\[\small PQ \parallel SR\]
Also from part (ii),\[\small PQ = SR\]
Thus, one pair of opposite sides of quadrilateral \(\small PQRS\) are equal and parallel.
Therefore,\[\small PQRS \text{ is a parallelogram}\]

Q.E.D.

🎯 Exam Significance Exam Significance ›

This problem develops strong understanding of the Midpoint Theorem, which is frequently used in board examinations.
Questions involving midpoint theorem and parallelogram properties are commonly asked in CBSE competency-based questions.
The logical chain of proving parallelism, equality and then identifying a parallelogram is highly useful for Olympiads and entrance exams.
This concept is also applied later in coordinate geometry and vector geometry.

📘 Concept & Theory Theory and Concepts Used ›

Midpoint Theorem: The segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of it.
In a rhombus, diagonals bisect each other at right angles.
If one angle of a parallelogram is \(90^\circ\), then the parallelogram becomes a rectangle.

🗺️ Solution Roadmap Step-by-step Plan ›

Use midpoint theorem in different triangles formed by diagonals \(\small AC\) and \(\small BD\).
Show that opposite sides of \(\small PQRS\) are equal and parallel.
Conclude that \(\small PQRS\) is a parallelogram.
Use the rhombus property \(\small AC \perp BD\) to prove that one angle of \(\small PQRS\) is \(\small 90^\circ\).
Hence conclude that \(\small PQRS\) is a rectangle.

📊 Graph / Figure Graph / Figure ›

Concept Diagram

Fig. 8.20-1

📐 Proof Proof ›

📌 Given

\(\small ABCD\) is a rhombus and \(\small P,\;Q,\;R,\;S\) are the mid-points of \(\small AB,\;BC,\;CD,\;DA\) respectively.

🎯 To Prove

\(\small PQRS \text{ is a rectangle}\)

✍️ Proof

Step-by-step Proof · 22 steps · 2 parts

Step 1: Proving that opposite sides are parallel and equal
Consider triangle \(\small ABD\).
Since \(\small P\) and \(\small S\) are the midpoints of \(\small AB\) and \(\small AD\) respectively,
by Midpoint Theorem,\[\small PS \parallel BD\] and \[\small PS=\frac{1}{2}BD \tag{1}\]
Now consider triangle \(\small CBD\).
Since \(\small Q\) and \(\small R\) are the midpoints of \(\small BC\) and \(\small CD\) respectively,
by Midpoint Theorem, \[\small QR \parallel BD\] and \[\small QR=\frac{1}{2}BD \tag{2}\]
From equations (1) and (2),\[\small PS = QR\] and \[\small PS \parallel QR\]
Now consider triangle \(\small ABC\).
Since \(\small P\) and \(\small Q\) are the midpoints of \(\small AB\) and \(\small BC\) respectively,
by Midpoint Theorem,\[\small PQ \parallel AC\] and \[\small PQ=\frac{1}{2}AC \tag{3}\]
In triangle \(\small ADC\), \(\small S\) and \(\small R\) are the midpoints of \(\small AD\) and \(\small CD\) respectively.
Therefore, by Midpoint Theorem,\[\small SR \parallel AC\] and \[\small SR=\frac{1}{2}AC \tag{4}\]
From equations (3) and (4),\[\small PQ = SR\] and \[\small PQ \parallel SR\]
Thus both pairs of opposite sides of quadrilateral \(\small PQRS\) are equal and parallel.
Therefore, \[\small PQRS \text{ is a parallelogram}\]
Step 2: Using the property of rhombus
Since \(ABCD\) is a rhombus, its diagonals are perpendicular.
Therefore,\[\small AC \perp BD\]
Also, \[\small PQ \parallel AC\] and \[\small PS \parallel BD\]
Since lines parallel to perpendicular lines are also perpendicular,\[\small PQ \perp PS\]
Therefore,\[\small \angle SPQ = 90^\circ\]
A parallelogram having one right angle is a rectangle.
Hence,\[\small PQRS \text{ is a rectangle}\]

Q.E.D.

🎯 Exam Significance Exam Significance ›

This question combines Midpoint Theorem, properties of rhombus and properties of parallelogram in a single proof.
Such logical geometry proofs are frequently asked in CBSE board examinations.
Competitive examinations often test the ability to identify hidden parallel lines and use diagonal properties effectively.
This problem strengthens analytical thinking and proof-writing skills needed for Olympiads and entrance examinations.

📘 Concept & Theory Theory and Concepts Used ›

Midpoint Theorem: The line segment joining the midpoints of two sides of a triangle is parallel to the third side and equal to half of the third side.
In a rectangle, diagonals are equal: \[ AC = BD \]
A parallelogram having all sides equal is called a rhombus.

🗺️ Solution Roadmap Step-by-step Plan ›

Use midpoint theorem in triangles formed by diagonals \(\small AC\) and \(\small BD\).
Show that opposite sides of \(\small PQRS\) are equal and parallel.
Conclude that \(\small PQRS\) is a parallelogram.
Use the rectangle property \(\small AC = BD\).
Show that all four sides of \(\small PQRS\) re equal and conclude that it is a rhombus.

📊 Graph / Figure Graph / Figure ›

Fig. 8.20.2

📐 Proof Proof ›

📌 Given

\(\small ABCD\) is a rectangle and \(\small P,\;Q,\;R,\;S\) are the mid-points of \(\small AB,\;BC,\;CD,\;DA\) respectively.

🎯 To Prove

\[\small PQRS \text{ is a rhombus}\]

✍️ Proof

Step-by-step Proof · 28 steps · 2 parts

Step 1: Showing opposite sides are equal and parallel
Consider triangle \(\small ADC\).
Since \(\small S\) and \(\small R\) are the mid-points of \(\small AD\) and \(\small CD\) respectively,
by the Midpoint Theorem,\[\small SR \parallel AC\] and \[\small SR=\frac{1}{2}AC \tag{1}\]
Now consider triangle \(\small ABC\).
Since \(\small P\) and \(\small Q\) are the mid-points of \(\small AB\) and \(\small BC\) respectively,
by the Midpoint Theorem,\[\small PQ \parallel AC\] and \[\small PQ=\frac{1}{2}AC \tag{2}\]
From equations (1) and (2),\[PQ = SR\] and \[\small PQ \parallel SR\]
Therefore, one pair of opposite sides of quadrilateral \(\small PQRS\) are equal and parallel.
Now consider triangle \(\small ABD\).
Since \(\small P\) and \(\small S\) are the mid-points of \(\small AB\) and \(\small AD\),
by Midpoint Theorem,\[\small PS \parallel BD\] and \[\small PS=\frac{1}{2}BD \tag{3}\]
Consider triangle \(CBD\).
Since \(\small Q\) and \(\small R\) are the mid-points of \(\small BC\) and \(\small CD\),
by Midpoint Theorem,\[\small QR \parallel BD\] and \[\small QR=\frac{1}{2}BD \tag{4}\]
From equations (3) and (4),\[\small PS = QR\] and \[\small PS \parallel QR\]
Therefore, the second pair of opposite sides of \(\small PQRS\) are also equal and parallel.
Hence,\[\small PQRS \text{ is a parallelogram}\]
Step 2: Using properties of a rectangle
In a rectangle, diagonals are equal.
Therefore,\[\small AC = BD\]
From equations (1) and (2),\[\small PQ = SR = \frac{1}{2}AC\]
From equations (3) and (4),\[\small PS = QR = \frac{1}{2}BD\]
Since, \[\small AC=BD\]
therefore, \[\small \frac{1}{2}AC = \frac{1}{2}BD\]
Hence,\[\small PQ = QR = RS = SP\]
Thus all four sides of parallelogram \(\small PQRS\) are equal.
A parallelogram having all sides equal is a rhombus.
Therefore,
\[\small PQRS \text{ is a rhombus}\]

Q.E.D.

🎯 Exam Significance Exam Significance ›

This question combines properties of rectangles, rhombuses and midpoint theorem into a single geometric proof.
Such theorem-based reasoning questions are commonly asked in CBSE board examinations.
Competitive entrance exams frequently test the ability to connect multiple geometric properties logically.
This problem also builds strong proof-writing and analytical geometry skills useful for Olympiads and higher mathematics.

📘 Concept & Theory Theory and Concepts Used ›

Converse of Midpoint Theorem: A line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side.
If two lines are parallel to the same line, then they are parallel to each other.
A trapezium is a quadrilateral having one pair of opposite sides parallel.

🗺️ Solution Roadmap Step-by-step Plan ›

Use the fact that \(\small EF \parallel AB\) and \(\small AB \parallel DC\) to conclude that \(\small EF \parallel DC\).
Consider triangle \(\small DBC\).
Show that \(\small E\) is the midpoint of side \(\small DB\) using the converse of midpoint theorem in triangle \(\small ABD\).
Apply midpoint theorem in triangle \(\small DBC\) to prove that \(\small F\) bisects \(\small BC\).

📊 Graph / Figure Graph / Figure ›

Fig. 1 — Free body diagram

📐 Proof Proof ›

📌 Given

\(\small ABCD\) is a trapezium such that \[\small AB \parallel DC\]

\(\small BD\) is a diagonal, \(E\) is the midpoint of \(\small AD\), and a line through \(\small E\) parallel to \(\small AB\) intersects \(\small BC\) at \(\small F\).

Therefore,

\[\small EF \parallel AB \]

🎯 To Prove

\[\small F \text{ is the midpoint of } BC\]

✍️ Proof

Step-by-step Proof · 13 steps · 2 parts

Step 1: Consider triangle \(\small ABD\)
In triangle \(\small ABD\), \(\small E\) is the midpoint of \(\small AD\).
Also, \[\small EF \parallel AB\]
The line through the midpoint of one side of a triangle and parallel to another side bisects the third side.
Therefore, in triangle \(\small ABD\), point \(\small F\) bisects side \(\small DB\).
Hence,\[\small DF = FB\]
thus, \[\small F\text{ is the midpoint of } DB.\]
Step 2: Consider triangle \(\small DBC\)
We are given that, \[\small AB \parallel DC\] and \[\small EF \parallel AB\]
Therefore,\[\small EF \parallel DC\]
Now consider triangle \(\small DBC\).
In this triangle, \(\small F\) is the midpoint of \(\small DB\) and the line through \(\small F\) is parallel to \(\small DC\).
By the converse of midpoint theorem, the line parallel to one side of a triangle through the midpoint of another side bisects the third side
Therefore,\[\small BF = FC\]
Hence,\[\small F \text{ is the midpoint of } BC\]

Q.E.D.

🎯 Exam Significance Exam Significance ›

This problem is an important application of the converse of midpoint theorem.
CBSE board examinations frequently ask geometry proofs involving parallel lines and midpoint concepts.
Competitive entrance examinations test the ability to identify hidden triangles and apply midpoint theorem strategically.
This question improves logical deduction and proof-writing skills essential for higher geometry.

📘 Concept & Theory Theory and Concepts Used ›

In a parallelogram, opposite sides are parallel: \[ AB \parallel CD \quad \text{and} \quad AD \parallel BC \]
Midpoint Theorem: The segment joining the midpoint of one side of a triangle and parallel to another side bisects the third side.
A diagonal is trisected when it is divided into three equal parts.

🗺️ Solution Roadmap Step-by-step Plan ›

Draw diagonal \(\small BD\) and let \(\small AF\) and \(\small EC\) intersect it at points \(\small P\) and \(\small Q\) respectively.
Use midpoint theorem in suitable triangles to prove that \(\small P\) and \(\small Q\) divide \(\small BD\) equally.
Show that \[\small DP = PQ = QB\]
Hence conclude that \(\small AF\) and \(\small EC\) trisect diagonal \(\small BD\).

📊 Graph / Figure Graph / Figure ›

Fig. 8.22

📐 Proof Proof ›

📌 Given

\(\small ABCD\) is a parallelogram.

\(\small E\) and \(\small F\) are the mid-points of \(\small AB\) and \(\small CD\) respectively.

Let

\[\small AF \cap BD = P \]

and

\[\small EC \cap BD = Q \]

🎯 To Prove

\[ DP = PQ = QB \]

Hence, \(AF\) and \(EC\) trisect diagonal \(BD\).

✍️ Proof

Step-by-step Proof · 24 steps · 4 parts

Step 1: Using midpoint theorem in triangle \(\small DAB\)
In parallelogram \(\small ABCD\),
\[\small AB \parallel CD\]
Given
Since \(\small F\) is the midpoint of \(\small CD\),\[\small CF = FD \]
Also,\[\small AB = CD\]
Therefore, \[\small CF = \frac{1}{2}CD = \frac{1}{2}AB\]
Since \(\small E\) is midpoint of \(\small AB\),\[\small AE = EB = \frac{1}{2}AB\]
Hence, \[\small AE = CF\]
Also, \[\small AE \parallel CF\]
Therefore quadrilateral \(\small AECF\) is a parallelogram.
Hence opposite sides are parallel:\[\small AF \parallel EC \tag{1}\]
Step 2: Showing that \(P\) trisects \(DQ\)
Consider triangle \(\small DCF\).
Since \(\small F\) is midpoint of \(\small DC\),\[\small DF = FC\]
Also from equation (1),\[\small FP \parallel EC\]
In triangle \(\small DQC\), point \(\small F\) is midpoint of \(\small DC\) and line through \(\small F\) parallel to \(\small QC\) meets \(\small DQ\) at \(\small P\).
By midpoint theorem,\[\small DP = PQ \tag{2}\]
Step 3: Showing that \(Q\) trisects \(PB\)
Consider triangle \(\small PAB\).
Since \(E\) is midpoint of \(AB\),\[\small AE = EB\]
Also from equation (1),\[\small EQ \parallel AP\]
In triangle \(\small PAB\), the line through midpoint \(\small E\) parallel to side \(\small AP\) bisects side \(\small PB\).
Therefore,\[\small PQ = QB \tag{3}\]
Step 4: Combining the results
From equations (2) and (3),\[\small DP = PQ\] and \[\small PQ = QB\]
Therefore,\[\small DP = PQ = QB\]
Thus diagonal \(\small BD\) is divided into three equal parts.
Hence,\[\small AF \text{ and } EC \text{ trisect diagonal } BD\]

Q.E.D.

🎯 Exam Significance Exam Significance ›

This question is an excellent application of midpoint theorem in a parallelogram.
It develops the ability to identify hidden triangles and parallel lines in geometry proofs.
Such multi-step proof questions are frequently asked in CBSE board examinations and Olympiads.
Competitive entrance examinations often test logical deduction using midpoint and parallelogram properties together.

Q6

NUMERIC3 marks

\(\small ABC\) is a triangle right angled at \(\small C\). A line through the mid-point \(\small M\) of hypotenuse \(\small AB\) and parallel to \(\small BC\) intersects \(\small AC\) at \(\small D\). Show that:

(i) \(\small D\) is the mid-point of \(\small AC\)
(ii) \(\small MD \perp AC\)
(iii) \(\small CM = MA = \frac{1}{2}AB\)

📘 Concept & Theory Theory and Concepts Used ›

If a line is drawn through the midpoint of one side of a triangle parallel to another side, then it bisects the third side.
If a line is perpendicular to one of two parallel lines, then it is perpendicular to the other.
In a right triangle, the midpoint of the hypotenuse is equidistant from all three vertices.

🗺️ Solution Roadmap Step-by-step Plan ›

Use midpoint theorem in triangle \(\small ABC\) with \(\small M\) as midpoint of hypotenuse \(\small AB\).
Show that \(\small D\) bisects side \(\small AC\).
Use parallel line property to prove \(\small MD \perp AC\).
Use the theorem related to midpoint of hypotenuse in a right triangle.

📊 Graph / Figure Graph / Figure ›

Fig. 8.22-1

📐 Proof Proof ›

📌 Given

\(ABC\) is a right-angled triangle such that

\[ \angle ACB = 90^\circ \]

\(M\) is the midpoint of hypotenuse \(AB\).

Through \(M\), a line parallel to \(BC\) intersects \(AC\) at \(D\).

Therefore,

\[ MD \parallel BC \]

🎯 To Prove

(i) \(\small D\) is the mid-point of \(\small AC\)
(ii) \(\small MD \perp AC\)
(iii) \(\small CM = MA = \frac{1}{2}AB\)

✍️ Proof

Step-by-step Proof · 14 steps · 3 parts

Part (i) - Proving that \(\small D\) is the midpoint of \(\small AC\)
In triangle \(\small ABC\), point \(\small M\) is the midpoint of side \(\small AB\).
Also, \[\small MD \parallel BC\]
A line drawn through the midpoint of one side of a triangle parallel to another side bisects the third side.
Therefore,\[\small AD = DC\]
Hence,\[\small D \text{ is the midpoint of } AC\]
Part (ii) - Proving that \(\small MD \perp AC\)
Since triangle \(\small ABC\) is right angled at \(\small C\),\[\small BC \perp AC\]
Also, \[\small MD \parallel BC\]
A line parallel to a line perpendicular to another line is also perpendicular to that line.
Therefore, \[\small MD \perp AC\]
Part (iii) Proving \(\small CM = MA = \frac{1}{2}AB\)
Since \(\small M\) is the midpoint of hypotenuse \(AB\),\[\small AM = MB = \frac{1}{2}AB\]
In a right triangle, the midpoint of the hypotenuse is equidistant from all three vertices
Therefore,\[\small MA = MB = MC\]
Since, \[\small MA = \frac{1}{2}AB\]
therefore,\[\small CM = MA = \frac{1}{2}AB\]

Q.E.D.

🎯 Exam Significance Exam Significance ›

This question is a very important application of midpoint theorem in right triangles.
The property of midpoint of hypotenuse is frequently asked in CBSE board examinations.
Competitive examinations often use this theorem to simplify geometry and coordinate geometry problems.
This problem strengthens understanding of parallel lines, perpendicularity and triangle geometry.

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