RBSE Solutions Class 9 Maths Chapter 9 Quadrilaterals Exercise 9.2

 

Question 1. In the given figure, if ABCD and AEFG are parallelograms, and \( \angle C = 55^\circ \), then find the value of \( \angle F \).

Answer: In parallelogram ABCD, the opposite angles are equal. So, if \( \angle C = 55^\circ \), then \( \angle A = 55^\circ \). Also, AEFG is a parallelogram. In a parallelogram, opposite angles are equal. This means \( \angle A = \angle F \). Since \( \angle A \) is \( 55^\circ \), then \( \angle F \) is also \( 55^\circ \).
In simple words: Since opposite angles in a parallelogram are equal, \( \angle A \) will be the same as \( \angle C \). And since \( \angle A \) is also opposite to \( \angle F \) in the second parallelogram, \( \angle F \) will be the same as \( \angle A \). So, \( \angle F \) is \( 55^\circ \).

🎯 Exam Tip: Remember that opposite angles in any parallelogram are always equal. This is a key property for solving such problems.

 

Question 2. Can all the angles of a quadrilateral be acute angles? Give reason for your answer.
Answer: No, all the angles of a quadrilateral cannot be acute. An acute angle is always less than \( 90^\circ \). If all four angles were acute, their sum would be less than \( 4 \times 90^\circ = 360^\circ \). However, the sum of all angles in any quadrilateral is exactly \( 360^\circ \). Therefore, it is impossible for all four angles to be acute.
In simple words: No, because if all four angles were smaller than \( 90^\circ \), their total would be less than \( 360^\circ \). But all angles in a four-sided shape must add up to exactly \( 360^\circ \).

🎯 Exam Tip: Always remember the angle sum property for different polygons: triangles (\( 180^\circ \)), quadrilaterals (\( 360^\circ \)), pentagons (\( 540^\circ \)), and so on. This is a fundamental concept.

 

Question 3. Can all the angles of a quadrilateral be right angles? Give reason for your answer.
Answer: Yes, all the angles of a quadrilateral can be right angles. A right angle is exactly \( 90^\circ \). If all four angles are \( 90^\circ \), their sum will be \( 4 \times 90^\circ = 360^\circ \). This matches the angle sum property of a quadrilateral, which is \( 360^\circ \). When all angles are right angles, the quadrilateral is either a square or a rectangle.
In simple words: Yes, a quadrilateral can have all right angles. If each angle is \( 90^\circ \), their total is \( 360^\circ \), which is what a four-sided shape needs. This kind of shape is a square or a rectangle.

🎯 Exam Tip: Understand that a square is a special type of rectangle, meaning all squares are rectangles, but not all rectangles are squares. Both have four right angles.

 

Question 4. Diagonals of a quadrilateral bisect each other. If \( \angle A = 35^\circ \) then find \( \angle B \).
Answer: If the diagonals of a quadrilateral bisect each other, it means the quadrilateral is a parallelogram. In a parallelogram, consecutive angles (angles next to each other) are supplementary, meaning they add up to \( 180^\circ \). So, \( \angle A + \angle B = 180^\circ \). Given that \( \angle A = 35^\circ \), we can find \( \angle B \):
\( 35^\circ + \angle B = 180^\circ \)
\( \implies \angle B = 180^\circ - 35^\circ \)
\( \implies \angle B = 145^\circ \)
In simple words: When a shape's diagonals cut each other in half, it's a parallelogram. In these shapes, angles next to each other add up to \( 180^\circ \). So, if \( \angle A \) is \( 35^\circ \), then \( \angle B \) must be \( 180^\circ - 35^\circ \), which is \( 145^\circ \).

🎯 Exam Tip: Remember the properties of parallelograms: opposite sides are parallel and equal, opposite angles are equal, and consecutive angles are supplementary. These are fundamental for quadrilateral problems.

 

Question 5. Opposite angles of a quadrilateral are equal. If AB = 4 cm then determine the length of CD.
Answer: If the opposite angles of a quadrilateral are equal, then the quadrilateral is a parallelogram. One of the important properties of a parallelogram is that its opposite sides are equal in length. Therefore, if the length of side AB is 4 cm, then the length of its opposite side CD will also be 4 cm.
\( CD = AB = 4 \text{ cm} \)
In simple words: A shape with equal opposite angles is a parallelogram. In a parallelogram, opposite sides are always the same length. So, if side AB is 4 cm, then side CD must also be 4 cm.

🎯 Exam Tip: Always relate the given properties of a quadrilateral (like equal opposite angles or bisecting diagonals) back to specific types of quadrilaterals (like parallelograms) to deduce other properties needed for the solution.

 

Question 6. ABCD is a rhombus in which an altitude drawn from D to AB bisects AB. Find the angles of a rhombus.
Answer: Let ABCD be a rhombus. Let the altitude from vertex D to side AB meet AB at point E. We are given that DE bisects AB, meaning E is the midpoint of AB.
In a rhombus, all sides are equal in length. So, \( AD = AB = BC = CD \).
Consider \( \triangle ADE \). It is a right-angled triangle at E.
Since DE is an altitude that bisects the side AB in \( \triangle ABD \), this implies that \( \triangle ABD \) is an isosceles triangle with \( AD = BD \).
We know that in a rhombus, \( AD = AB \).
Therefore, we have \( AD = AB = BD \). This means \( \triangle ABD \) is an equilateral triangle.
In an equilateral triangle, all angles are \( 60^\circ \). So, \( \angle A = 60^\circ \).
In a rhombus, consecutive angles are supplementary (add up to \( 180^\circ \)).
So, \( \angle A + \angle B = 180^\circ \) and \( \angle A + \angle D = 180^\circ \).
\( 60^\circ + \angle B = 180^\circ \implies \angle B = 120^\circ \).
\( 60^\circ + \angle D = 180^\circ \implies \angle D = 120^\circ \).
Also, opposite angles in a rhombus are equal, so \( \angle C = \angle A = 60^\circ \).
Thus, the angles of the rhombus are \( 60^\circ, 120^\circ, 60^\circ, 120^\circ \).

In simple words: When the height from D cuts AB exactly in half, it means that triangle ABD is a special triangle where all three sides are equal (equilateral). This makes angle A equal to \( 60^\circ \). In a rhombus, angles next to each other add up to \( 180^\circ \). So, angle D will be \( 180^\circ - 60^\circ = 120^\circ \). The angles of the rhombus are then \( 60^\circ, 120^\circ, 60^\circ, 120^\circ \).

🎯 Exam Tip: Recognize that if an altitude from a vertex of a triangle also bisects the opposite side, that triangle must be isosceles. If it's a rhombus, this often leads to an equilateral triangle.

 

Question 8. D, E and F are the mid-points of the sides BC, CA and AB respectively of an equilateral \( \triangle ABC \). Show that \( \triangle DEF \) is also an equilateral triangle.
Answer: We are given that D, E, and F are the midpoints of sides BC, CA, and AB respectively of \( \triangle ABC \). We assume that \( \triangle ABC \) is an equilateral triangle, as implied by the question asking to prove \( \triangle DEF \) is also equilateral. In an equilateral triangle, all three sides are equal: \( AB = BC = CA \).
By the Midpoint Theorem:
1. The line segment joining the midpoints of two sides of a triangle is parallel to the third side and half its length.
Applying this:
- For side DE (joining midpoints of BC and CA), \( DE = \frac{1}{2} AB \).
- For side EF (joining midpoints of CA and AB), \( EF = \frac{1}{2} BC \).
- For side FD (joining midpoints of AB and BC), \( FD = \frac{1}{2} CA \).
Since \( AB = BC = CA \) (because \( \triangle ABC \) is equilateral), it follows that:
\( DE = EF = FD = \frac{1}{2} AB = \frac{1}{2} BC = \frac{1}{2} CA \).
Since all three sides of \( \triangle DEF \) are equal, \( \triangle DEF \) is also an equilateral triangle.
In simple words: When you connect the middle points of the sides of a triangle, the new triangle formed will have sides that are half the length of the original triangle's sides. If the first triangle has all equal sides (equilateral), then the new triangle will also have all equal sides, making it equilateral too.

🎯 Exam Tip: The Midpoint Theorem is crucial here. Remember that the segment connecting midpoints is not just parallel, but also exactly half the length of the third side. This theorem is often used to prove properties of smaller triangles formed within larger ones.

 

Question 9. The points P and Q have been taken on opposite sides AB and CD respectively of a parallelogram ABCD in such a way that AP = CQ (see figure). Show that AC and PQ bisect each other.
Answer: We are given a parallelogram ABCD. Points P and Q are on sides AB and CD respectively, such that \( AP = CQ \). We need to show that the diagonals AC and PQ bisect each other.

**Proof:**
1. Since ABCD is a parallelogram, its opposite sides are parallel. So, \( AB \parallel CD \).
2. As P is a point on AB and Q is a point on CD, it implies that \( AP \parallel CQ \).
3. We are given that \( AP = CQ \).
4. In a quadrilateral, if one pair of opposite sides is both parallel and equal, then the quadrilateral is a parallelogram.
5. Therefore, quadrilateral APCQ is a parallelogram.
6. The diagonals of a parallelogram bisect each other. The diagonals of parallelogram APCQ are AC and PQ.
7. Hence, AC and PQ bisect each other. This completes the proof.
In simple words: Because ABCD is a parallelogram, the line AP is parallel to CQ. We are also told that AP and CQ are the same length. If a four-sided shape has two sides that are both parallel and equal in length, it is a parallelogram. Since APCQ is a parallelogram, its diagonals (AC and PQ) must cut each other exactly in half.

🎯 Exam Tip: To prove that two line segments bisect each other, a common strategy is to show that they are the diagonals of a parallelogram. Mastering the conditions for a quadrilateral to be a parallelogram is key.

 

Question 10. E is the mid-point of the side AD of a trapezium ABCD in which \( AB \parallel DC \). Through E, a line is drawn parallel to AB which intersects BC at F. Show that F will be the mid-point of side BC.
Answer: We are given a trapezium ABCD where \( AB \parallel DC \). E is the midpoint of side AD. A line is drawn through E parallel to AB, and this line intersects BC at F. We need to prove that F is the midpoint of BC.
**Construction:** Draw a diagonal AC, which intersects EF at point O.

**Proof:**
1. We are given that \( EF \parallel AB \). Since \( AB \parallel DC \), it follows that \( EF \parallel DC \). Thus, \( EO \parallel DC \) (as O lies on EF).
2. Consider \( \triangle ADC \). E is the midpoint of AD (given). EO is parallel to DC (from step 1).
3. By the converse of the Midpoint Theorem (also known as the Intercept Theorem), if a line passes through the midpoint of one side of a triangle and is parallel to another side, then it bisects the third side. Therefore, O must be the midpoint of AC.
4. Now consider \( \triangle ABC \). O is the midpoint of AC (from step 3). OF is parallel to AB (since \( EF \parallel AB \), and O, F are on EF).
5. Again, by the converse of the Midpoint Theorem, F must be the midpoint of BC.
This proves that F is the midpoint of side BC.
In simple words: We draw a diagonal AC. In the triangle ADC, E is the middle of AD, and the line EO is parallel to DC. This means O must be the middle of AC. Now, in the triangle ABC, O is the middle of AC, and the line OF is parallel to AB. This means F must be the middle of BC.

🎯 Exam Tip: When dealing with trapeziums and midpoints, constructing a diagonal often helps to apply the Midpoint Theorem or its converse twice, for two different triangles.

 

Question 11. In \( \triangle ABC \), AB = 5 cm, BC = 8 cm and CA = 7 cm. If D and E are the mid-points of sides AB and BC respectively, then find the length of side DE.
Answer: We are given a triangle ABC with side lengths \( AB = 5 \text{ cm} \), \( BC = 8 \text{ cm} \), and \( CA = 7 \text{ cm} \). D is the midpoint of side AB, and E is the midpoint of side BC. We need to find the length of the line segment DE.
According to the Midpoint Theorem, the line segment joining the midpoints of two sides of a triangle is parallel to the third side and is half the length of the third side.
Here, D and E are the midpoints of AB and BC respectively. The third side is AC.
Therefore, \( DE = \frac{1}{2} AC \).
Given \( AC = 7 \text{ cm} \).
\( DE = \frac{1}{2} \times 7 \text{ cm} \)
\( \implies DE = 3.5 \text{ cm} \)

In simple words: The Midpoint Theorem tells us that if you connect the middle points of two sides of a triangle, that new line will be half the length of the third side. Here, D and E are midpoints of AB and BC, so the line DE is half the length of AC. Since AC is 7 cm, DE is 3.5 cm.

🎯 Exam Tip: Always identify the two midpoints and the third side correctly to apply the Midpoint Theorem. It's a direct application problem.

 

Question 13. In the figure, D, E and F are the mid-points of the sides BC, CA and AB respectively. If AB = 4.3 cm, BC = 5.6 cm and AC = 3.9 cm, then find the perimeter of the following:
(i) \( \triangle DEF \)
(ii) quad. BDEF

Answer: We are given \( \triangle ABC \) with side lengths \( AB = 4.3 \text{ cm} \), \( BC = 5.6 \text{ cm} \), and \( AC = 3.9 \text{ cm} \). D, E, and F are the midpoints of sides BC, CA, and AB respectively.
According to the Midpoint Theorem, the line segment connecting the midpoints of two sides of a triangle is parallel to the third side and is half its length.
Therefore, we can find the lengths of the sides of \( \triangle DEF \):
- Side DE connects midpoints of BC and CA. Thus, \( DE = \frac{1}{2} AB = \frac{1}{2} \times 4.3 \text{ cm} = 2.15 \text{ cm} \).
- Side EF connects midpoints of CA and AB. Thus, \( EF = \frac{1}{2} BC = \frac{1}{2} \times 5.6 \text{ cm} = 2.8 \text{ cm} \).
- Side FD connects midpoints of AB and BC. Thus, \( FD = \frac{1}{2} AC = \frac{1}{2} \times 3.9 \text{ cm} = 1.95 \text{ cm} \).

(i) **Perimeter of \( \triangle DEF \):**
The perimeter is the sum of its sides:
\( \text{Perimeter of } \triangle DEF = DE + EF + FD \)
\( = 2.15 \text{ cm} + 2.8 \text{ cm} + 1.95 \text{ cm} \)
\( = 6.9 \text{ cm} \)

(ii) **Perimeter of quadrilateral BDEF:**
The sides of quadrilateral BDEF are BD, DE, EF, and FB.
- D is the midpoint of BC, so \( BD = \frac{1}{2} BC = \frac{1}{2} \times 5.6 \text{ cm} = 2.8 \text{ cm} \).
- F is the midpoint of AB, so \( FB = \frac{1}{2} AB = \frac{1}{2} \times 4.3 \text{ cm} = 2.15 \text{ cm} \).
- We already found \( DE = 2.15 \text{ cm} \) and \( EF = 2.8 \text{ cm} \).
The perimeter is the sum of its sides:
\( \text{Perimeter of BDEF} = BD + DE + EF + FB \)
\( = 2.8 \text{ cm} + 2.15 \text{ cm} + 2.8 \text{ cm} + 2.15 \text{ cm} \)
\( = 9.9 \text{ cm} \)
In simple words: We use the Midpoint Theorem to find the lengths of the lines connecting the midpoints. Each line is half the length of the opposite side. Once we have these lengths, we just add them up to find the perimeter of the triangle DEF and the quadrilateral BDEF.

🎯 Exam Tip: When calculating perimeters of shapes formed by midpoints, always clearly identify all the sides of the shape and apply the Midpoint Theorem carefully to find their lengths. Avoid common mistakes in identifying which side corresponds to which midpoint connection.

 

Question 14. Show that the line segment joining the mid-points of the consecutive sides of a square is also a square.
Answer: Let ABCD be a square, and let E, F, G, H be the midpoints of the sides AB, BC, CD, and DA respectively. We need to prove that the quadrilateral EFGH, formed by joining these midpoints, is also a square.
**Proof:**
1. **Properties of a square:** In a square ABCD, all sides are equal (\( AB = BC = CD = DA \)) and all angles are \( 90^\circ \). Also, its diagonals are equal (\( AC = BD \)) and bisect each other at right angles.
2. **Using the Midpoint Theorem:**
- In \( \triangle ABC \), E and F are midpoints of AB and BC. By the Midpoint Theorem, \( EF = \frac{1}{2} AC \) and \( EF \parallel AC \).
- In \( \triangle ADC \), G and H are midpoints of CD and DA. By the Midpoint Theorem, \( HG = \frac{1}{2} AC \) and \( HG \parallel AC \).
- These two imply that \( EF = HG \) and \( EF \parallel HG \). So, EFGH is a parallelogram.
3. **Showing all sides are equal:**
- Similarly, considering the other diagonal BD:
- In \( \triangle ABD \), H and E are midpoints of AD and AB. By the Midpoint Theorem, \( HE = \frac{1}{2} BD \) and \( HE \parallel BD \).
- In \( \triangle BCD \), F and G are midpoints of BC and CD. By the Midpoint Theorem, \( FG = \frac{1}{2} BD \) and \( FG \parallel BD \).
- Since ABCD is a square, its diagonals are equal, so \( AC = BD \).
- Therefore, \( EF = HG = HE = FG \) (all being half of equal diagonals). This means EFGH is a rhombus (a parallelogram with all sides equal).
4. **Showing one angle is \( 90^\circ \):**
- Consider \( \triangle AEH \). Since E and H are midpoints of equal sides AB and AD (of square ABCD), \( AE = AH = \frac{1}{2} \text{side of square} \).
- Also, \( \angle A = 90^\circ \). So, \( \triangle AEH \) is an isosceles right-angled triangle.
- This means \( \angle AHE = \angle AEH = 45^\circ \).
- Similarly, consider \( \triangle DGH \). G and H are midpoints of equal sides CD and DA. So \( DG = DH = \frac{1}{2} \text{side of square} \).
- Also, \( \angle D = 90^\circ \). So, \( \triangle DGH \) is an isosceles right-angled triangle.
- This means \( \angle DHG = \angle DGH = 45^\circ \).
- Now, look at the straight line AD at point H. The sum of angles around H is \( 180^\circ \).
\( \angle AHE + \angle EHG + \angle DHG = 180^\circ \)
\( 45^\circ + \angle EHG + 45^\circ = 180^\circ \)
\( 90^\circ + \angle EHG = 180^\circ \)
\( \implies \angle EHG = 90^\circ \)
5. **Conclusion:** Since EFGH is a rhombus with one angle equal to \( 90^\circ \), it is a square.

In simple words: When you connect the middle points of the sides of a square, you create a new shape inside. Because the original shape is a square, all its sides are equal and its corners are \( 90^\circ \). Using these facts and the Midpoint Theorem, we can show that the new inner shape also has all equal sides and all \( 90^\circ \) corners, which means it is also a square.

🎯 Exam Tip: To prove a quadrilateral is a square, you usually need to show two things: first, that all its sides are equal (making it a rhombus), and second, that at least one of its angles is \( 90^\circ \) (which then makes all angles \( 90^\circ \)).

 

Question 15. The diagonals of a quadrilateral are perpendicular to each other. Show that the quadrilateral, formed by joining the mid-points of its sides, is a rectangle.
Answer: Let ABCD be any quadrilateral whose diagonals AC and BD are perpendicular to each other. Let P, Q, R, S be the midpoints of sides AB, BC, CD, and DA respectively. We need to prove that the quadrilateral PQRS is a rectangle.
**Proof:**
1. **Forming a parallelogram:**
- In \( \triangle ABC \), P and Q are the midpoints of AB and BC. By the Midpoint Theorem, \( PQ \parallel AC \) and \( PQ = \frac{1}{2} AC \).
- In \( \triangle ADC \), R and S are the midpoints of CD and DA. By the Midpoint Theorem, \( RS \parallel AC \) and \( RS = \frac{1}{2} AC \).
- From these two facts, we have \( PQ \parallel RS \) and \( PQ = RS \).
- Similarly, in \( \triangle ABD \), P and S are the midpoints of AB and DA. By the Midpoint Theorem, \( PS \parallel BD \) and \( PS = \frac{1}{2} BD \).
- In \( \triangle BCD \), Q and R are the midpoints of BC and CD. By the Midpoint Theorem, \( QR \parallel BD \) and \( QR = \frac{1}{2} BD \).
- From these two facts, we have \( PS \parallel QR \) and \( PS = QR \).
- Since both pairs of opposite sides of PQRS are parallel and equal, PQRS is a parallelogram.
2. **Showing an angle is \( 90^\circ \):**
- We are given that the diagonals of quadrilateral ABCD are perpendicular, i.e., \( AC \perp BD \).
- We know that \( PQ \parallel AC \) and \( PS \parallel BD \).
- If two lines are perpendicular, then any line parallel to one of them is perpendicular to any line parallel to the other.
- Therefore, \( PS \perp PQ \), which means \( \angle SPQ = 90^\circ \).
3. **Conclusion:** A parallelogram with one right angle is a rectangle. Since PQRS is a parallelogram with \( \angle SPQ = 90^\circ \), it is a rectangle.

In simple words: When you connect the middle points of the sides of any four-sided shape whose main diagonals cross at a right angle, you create a parallelogram. Because the original diagonals are perpendicular, the sides of this new parallelogram will also meet at right angles. This makes the new shape a rectangle.

🎯 Exam Tip: Remember that a quadrilateral formed by joining the midpoints of any quadrilateral is always a parallelogram. To make it a rectangle, the diagonals of the original quadrilateral must be perpendicular. To make it a rhombus, the diagonals of the original quadrilateral must be equal. To make it a square, the diagonals of the original quadrilateral must be both equal and perpendicular.

 

Question 16. Prove that in a right-angled triangle, the median bisecting the hypotenuse is half of the hypotenuse.
Answer: Let \( \triangle ABC \) be a right-angled triangle, with the right angle at B. Let BD be the median to the hypotenuse AC, which means D is the midpoint of AC. We need to prove that \( BD = \frac{1}{2} AC \).
**Construction:** Extend BD to a point E such that \( BD = DE \). Join CE.

**Proof:**
1. Consider \( \triangle ADB \) and \( \triangle CDE \):
- \( AD = DC \) (because D is the midpoint of AC)
- \( \angle ADB = \angle CDE \) (vertically opposite angles)
- \( BD = DE \) (by construction)
2. By the SAS (Side-Angle-Side) congruence criterion, \( \triangle ADB \cong \triangle CDE \).
3. Since the triangles are congruent, their corresponding parts are equal:
- \( AB = CE \)
- \( \angle DAB = \angle DCE \). These are alternate interior angles, which means \( AB \parallel CE \).
4. Now consider quadrilateral ABCE. We have \( AB \parallel CE \) and \( AB = CE \). Thus, ABCE is a parallelogram.
5. We are given that \( \angle ABC = 90^\circ \) (right-angled at B).
6. A parallelogram with one right angle is a rectangle. Therefore, ABCE is a rectangle.
7. In a rectangle, the diagonals are equal in length. So, \( AC = BE \).
8. From our construction, \( BE = BD + DE \). Since we made \( BD = DE \), we have \( BE = BD + BD = 2BD \).
9. Substituting this into \( AC = BE \), we get \( AC = 2BD \).
10. Rearranging this, we find \( BD = \frac{1}{2} AC \).
Thus, the median bisecting the hypotenuse is half of the hypotenuse.
In simple words: If you draw a line from the right angle corner of a triangle to the middle of the longest side (hypotenuse), that line will be exactly half the length of the longest side. You can prove this by extending that median to form a rectangle, and in a rectangle, the diagonals are equal.

🎯 Exam Tip: This is a very important theorem. The proof often involves extending the median to construct a parallelogram or a rectangle, which allows you to use properties of these quadrilaterals to show the relationship between the median and the hypotenuse.

 

Question 17. Prove that the quadrilateral formed by joining the mid-points of the consecutive sides of a rectangle is a rhombus.
Answer: Let ABCD be a rectangle. Let P, Q, R, and S be the midpoints of the sides AB, BC, CD, and DA respectively. We need to prove that the quadrilateral PQRS is a rhombus.
**Proof:**
1. **PQRS is a parallelogram:**
- In \( \triangle ABC \), P and Q are the midpoints of AB and BC. By the Midpoint Theorem, \( PQ \parallel AC \) and \( PQ = \frac{1}{2} AC \).
- In \( \triangle ADC \), R and S are the midpoints of CD and DA. By the Midpoint Theorem, \( RS \parallel AC \) and \( RS = \frac{1}{2} AC \).
- This means \( PQ \parallel RS \) and \( PQ = RS \).
- Similarly, in \( \triangle ABD \), P and S are the midpoints of AB and DA. By the Midpoint Theorem, \( PS \parallel BD \) and \( PS = \frac{1}{2} BD \).
- In \( \triangle BCD \), Q and R are the midpoints of BC and CD. By the Midpoint Theorem, \( QR \parallel BD \) and \( QR = \frac{1}{2} BD \).
- This means \( PS \parallel QR \) and \( PS = QR \).
- Since both pairs of opposite sides of PQRS are parallel and equal, PQRS is a parallelogram.
2. **Showing adjacent sides are equal:**
- In a rectangle ABCD, the diagonals are equal in length, so \( AC = BD \).
- We know \( PQ = \frac{1}{2} AC \) and \( PS = \frac{1}{2} BD \).
- Since \( AC = BD \), it follows that \( PQ = PS \).
- Alternatively, consider \( \triangle APS \) and \( \triangle BPQ \):
- \( AP = PB \) (P is the midpoint of AB)
- \( AS = BQ \) (Since AD = BC for a rectangle, and S and Q are midpoints, \( \frac{1}{2} AD = \frac{1}{2} BC \))
- \( \angle A = \angle B = 90^\circ \) (angles of a rectangle)
- So, \( \triangle APS \cong \triangle BPQ \) by SAS congruence criterion.
- This means \( PS = PQ \) (Corresponding Parts of Congruent Triangles are equal).
3. **Conclusion:** Since PQRS is a parallelogram and its adjacent sides (e.g., PQ and PS) are equal, it is a rhombus.

In simple words: When you connect the middle points of the sides of a rectangle, the new shape formed inside is a parallelogram. Since the diagonals of the original rectangle are equal, the sides of this parallelogram will also be equal to each other. A parallelogram with all equal sides is called a rhombus.

🎯 Exam Tip: To prove a quadrilateral formed by midpoints is a rhombus, you typically first prove it's a parallelogram, then show that its adjacent sides are equal. This often uses either the equality of the original quadrilateral's diagonals or the congruence of corner triangles.

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