Study Materials Available

Access summaries, videos, slides, infographics, mind maps and more

QUADRILATERALS - Q&A

Exercise 8.1

1. If the diagonals of a parallelogram are equal, then show that it is a rectangle.

Given: A parallelogram ABCD where diagonals AC = BD.
To Prove: ABCD is a rectangle.

Proof:
Consider triangles Δ ABC and Δ DCB.
1. AB = DC (Opposite sides of a parallelogram are equal)
2. BC = BC (Common side)
3. AC = BD (Given)

Therefore, by SSS congruence rule:
Δ ABC ≅ Δ DCB

Since the triangles are congruent, their corresponding angles are equal:
∠ ABC = ∠ DCB (by CPCT)

We know that consecutive interior angles of a parallelogram sum to 180° (since AB || DC):
∠ ABC + ∠ DCB = 180°

Substituting ∠ DCB with ∠ ABC:
∠ ABC + ∠ ABC = 180°
2 ∠ ABC = 180°
∠ ABC = 90°

Since one angle of the parallelogram is 90°, ABCD is a rectangle.

2. Show that the diagonals of a square are equal and bisect each other at right angles.

Given: A square ABCD.
To Prove: (i) AC = BD, (ii) AC bisects BD, (iii) AC ⊥ BD.

(i) Prove Diagonals are Equal:
In Δ ABC and Δ BAD:
1. AB = AB (Common)
2. BC = AD (Sides of a square are equal)
3. ∠ ABC = ∠ BAD = 90° (Angles of a square are 90°)
So, Δ ABC ≅ Δ BAD (by SAS).
Therefore, AC = BD (by CPCT).

(ii) Prove Diagonals Bisect Each Other:
Since a square is a parallelogram, its diagonals bisect each other. So, OA = OC and OB = OD.

(iii) Prove Diagonals are Perpendicular:
In Δ AOB and Δ AOD:
1. AO = AO (Common)
2. AB = AD (Sides of a square)
3. OB = OD (Diagonals bisect each other)
So, Δ AOB ≅ Δ AOD (by SSS).

Therefore, ∠ AOB = ∠ AOD (by CPCT).
But ∠ AOB + ∠ AOD = 180° (Linear Pair).
So, 2 ∠ AOB = 180° ⇒ ∠ AOB = 90°.
Hence, diagonals bisect each other at right angles.

3. Diagonal AC of a parallelogram ABCD bisects ∠ A (see Fig. 8.11). Show that
(i) it bisects ∠ C also,
(ii) ABCD is a rhombus.

Given: Parallelogram ABCD, AC bisects ∠ A (so ∠ DAC = ∠ BAC).

(i) To Prove: AC bisects ∠ C.
Since ABCD is a parallelogram, AD || BC and AC is a transversal.
∠ DAC = ∠ BCA (Alternate Interior Angles) ... (1)
Similarly, AB || DC and AC is a transversal.
∠ BAC = ∠ DCA (Alternate Interior Angles) ... (2)

Given that ∠ DAC = ∠ BAC.
From (1) and (2), we get:
∠ BCA = ∠ DCA.
Therefore, AC bisects ∠ C.

(ii) To Prove: ABCD is a rhombus.
From (i), we have ∠ DAC = ∠ BCA.
But we also know ∠ DAC = ∠ BAC (Given).
Therefore, ∠ BAC = ∠ BCA.
In Δ ABC, sides opposite to equal angles are equal.
So, AB = BC.
Since adjacent sides of a parallelogram are equal (AB = BC), all sides are equal (AB = BC = CD = DA).
Hence, ABCD is a rhombus.

4. ABCD is a rectangle in which diagonal AC bisects ∠ A as well as ∠ C. Show that:
(i) ABCD is a square
(ii) diagonal BD bisects ∠ B as well as ∠ D.

(i) To Prove: ABCD is a square.
Given AC bisects ∠ A, so ∠ DAC = ∠ BAC.
Since ABCD is a rectangle, ∠ A = 90°.
So, ∠ DAC = ∠ BAC = 45°.
Similarly, AC bisects ∠ C, so ∠ DCA = ∠ BCA = 45°.

In Δ ABC, ∠ BAC = ∠ BCA = 45°.
Sides opposite to equal angles are equal, so AB = BC.
Since adjacent sides of a rectangle are equal, it becomes a square.

(ii) To Prove: BD bisects ∠ B and ∠ D.
Since ABCD is a square (proved above), diagonals of a square bisect the angles.
Consider Δ BCD. BC = CD, so ∠ CBD = ∠ CDB.
Also ∠ C = 90°, so ∠ CBD + ∠ CDB = 90°.
Therefore, ∠ CBD = ∠ CDB = 45°.
Since ∠ B = 90°, ∠ ABD = 90° - 45° = 45°.
So, ∠ ABD = ∠ CBD, meaning BD bisects ∠ B.
Similarly, BD bisects ∠ D.

5. In parallelogram ABCD, two points P and Q are taken on diagonal BD such that DP = BQ (see Fig. 8.12). Show that:
(i) Δ APD ≅ Δ CQB
(ii) AP = CQ
(iii) Δ AQB ≅ Δ CPD
(iv) AQ = CP
(v) APCQ is a parallelogram

(i) In Δ APD and Δ CQB:
1. AD = CB (Opposite sides of parallelogram)
2. ∠ ADP = ∠ CBQ (Alternate interior angles, since AD || BC)
3. DP = BQ (Given)
So, Δ APD ≅ Δ CQB (by SAS).

(ii) Since Δ APD ≅ Δ CQB:
AP = CQ (by CPCT).

(iii) In Δ AQB and Δ CPD:
1. AB = CD (Opposite sides of parallelogram)
2. ∠ ABQ = ∠ CDP (Alternate interior angles, since AB || DC)
3. BQ = DP (Given)
So, Δ AQB ≅ Δ CPD (by SAS).

(iv) Since Δ AQB ≅ Δ CPD:
AQ = CP (by CPCT).

(v) From (ii) and (iv), we have:
AP = CQ and AQ = CP.
Since opposite sides of quadrilateral APCQ are equal, APCQ is a parallelogram.

6. ABCD is a trapezium in which AB || CD and AD = BC (see Fig. 8.13). Show that
(i) ∠ A = ∠ B
(ii) ∠ C = ∠ D
(iii) Δ ABC ≅ Δ BAD
(iv) diagonal AC = diagonal BD

[Hint: Extend AB and draw a line through C parallel to DA intersecting AB produced at E.]

(i) Construction: Draw CE || DA meeting AB produced at E.
Since AB || CD, AE || DC. Also AD || CE (Construction).
So, ADCE is a parallelogram.
Therefore, AD = CE (Opposite sides).
Given AD = BC, so BC = CE.
In Δ BCE, since BC = CE, ∠ CBE = ∠ CEB.
Also, ∠ A + ∠ CEB = 180° (Consecutive interior angles, AD || CE).
∠ B + ∠ CBE = 180° (Linear Pair).
So, ∠ A = ∠ B.

(ii) ∠ A + ∠ D = 180° (Consecutive interior angles).
∠ B + ∠ C = 180° (Consecutive interior angles).
Since ∠ A = ∠ B, then 180° - ∠ D = 180° - ∠ C.
So, ∠ C = ∠ D.

(iii) In Δ ABC and Δ BAD:
1. AB = BA (Common)
2. BC = AD (Given)
3. ∠ B = ∠ A (Proved in i)
So, Δ ABC ≅ Δ BAD (by SAS).

(iv) Since Δ ABC ≅ Δ BAD:
AC = BD (by CPCT).

Exercise 8.2

1. ABCD is a quadrilateral in which P, Q, R and S are mid-points of the sides AB, BC, CD and DA (see Fig. 8.20). AC is a diagonal. Show that:
(i) SR || AC and SR = 1/2 AC
(ii) PQ = SR
(iii) PQRS is a parallelogram.

(i) In Δ ADC:
S is the mid-point of DA and R is the mid-point of DC.
By Mid-point Theorem, the line segment joining the mid-points of two sides of a triangle is parallel to the third side and half of it.
So, SR || AC and SR = 1/2 AC.

(ii) Similarly, in Δ ABC:
P is the mid-point of AB and Q is the mid-point of BC.
By Mid-point Theorem, PQ || AC and PQ = 1/2 AC.
From (i), SR = 1/2 AC.
Therefore, PQ = SR.

(iii) We have proved PQ || AC and SR || AC, so PQ || SR.
Also PQ = SR.
Since a pair of opposite sides (PQ and SR) are equal and parallel, PQRS is a parallelogram.

2. ABCD is a rhombus and P, Q, R and S are the mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rectangle.

Step 1: PQRS is a parallelogram.
(Using the logic from Q1, joining midpoints of any quadrilateral forms a parallelogram).
PQ || AC and PQ = 1/2 AC.
SR || AC and SR = 1/2 AC.
So PQRS is a parallelogram.

Step 2: Prove one angle is 90°.
Diagonals of a rhombus intersect at 90°. Let diagonals AC and BD intersect at O.
Let PQ intersect AC at X and QR intersect BD at Y.
Since PQ || AC, QX || OY.
Since QR || BD (By midpoint theorem in Δ CBD), QY || OX.
Therefore, OXQY is a parallelogram.
Opposite angles of a parallelogram are equal.
∠ XOY = 90° (Diagonals of rhombus).
So, ∠ XQY = 90° (i.e., ∠ PQR = 90°).
Since PQRS is a parallelogram with one right angle, it is a rectangle.

3. ABCD is a rectangle and P, Q, R and S are mid-points of the sides AB, BC, CD and DA respectively. Show that the quadrilateral PQRS is a rhombus.

Step 1: PQRS is a parallelogram (by Mid-point theorem as in Q1).
So, PQ || AC and PQ = 1/2 AC.
And QR || BD and QR = 1/2 BD.

Step 2: Prove adjacent sides are equal.
In a rectangle, diagonals are equal. So AC = BD.
Therefore, 1/2 AC = 1/2 BD.
This implies PQ = QR.
Since adjacent sides of parallelogram PQRS are equal, PQRS is a rhombus.

4. ABCD is a trapezium in which AB || DC, BD is a diagonal and E is the mid-point of AD. A line is drawn through E parallel to AB intersecting BC at F (see Fig. 8.21). Show that F is the mid-point of BC.

Let the line EF intersect diagonal BD at G.

Step 1: In Δ DAB:
E is the mid-point of AD and EG || AB (Since EF || AB).
By Converse of Mid-point Theorem, G must be the mid-point of BD.

Step 2: In Δ BDC:
G is the mid-point of BD (Proved above).
GF || DC (Since EF || AB and AB || DC, so EF || DC).
By Converse of Mid-point Theorem, F is the mid-point of BC.
Hence Proved.

5. In a parallelogram ABCD, E and F are the mid-points of sides AB and CD respectively (see Fig. 8.22). Show that the line segments AF and EC trisect the diagonal BD.

To Prove: DP = PQ = QB.

Step 1: Show AECF is a parallelogram.
AB || CD ⇒ AE || CF.
AE = 1/2 AB and CF = 1/2 CD.
Since AB = CD (Opposite sides), AE = CF.
Since AE || CF and AE = CF, AECF is a parallelogram.
So, AF || EC (Opposite sides).

Step 2: In Δ DQC:
F is the mid-point of DC and FP || CQ (Since AF || EC).
By Converse of Mid-point Theorem, P is the mid-point of DQ.
So, DP = PQ ... (1)

Step 3: In Δ ABP:
E is the mid-point of AB and EQ || AP (Since EC || AF).
By Converse of Mid-point Theorem, Q is the mid-point of BP.
So, PQ = QB ... (2)

From (1) and (2):
DP = PQ = QB.
Hence, AF and EC trisect the diagonal BD.

6. ABC is a triangle right angled at C. A line through the mid-point M of hypotenuse AB and parallel to BC intersects AC at D. Show that
(i) D is the mid-point of AC
(ii) MD ⊥ AC
(iii) CM = MA = 1/2 AB

(i) In Δ ABC:
M is the mid-point of AB and MD || BC.
By Converse of Mid-point Theorem, D is the mid-point of AC.

(ii) Since MD || BC and AC is transversal:
∠ ADM = ∠ ACB (Corresponding angles).
Since ∠ ACB = 90°, ∠ ADM = 90°.
Therefore, MD ⊥ AC.

(iii) Join MC.
In Δ ADM and Δ CDM:
1. AD = CD (D is mid-point)
2. ∠ ADM = ∠ CDM = 90°
3. DM = DM (Common)
So, Δ ADM ≅ Δ CDM (by SAS).
Therefore, MA = MC (by CPCT).
We know MA = 1/2 AB (M is mid-point).
So, CM = MA = 1/2 AB.

Quick Navigation:

Quick Review Flashcards - Click to flip and test your knowledge!

Question

What three geometric components does every quadrilateral possess four of?

Answer

Every quadrilateral has four sides, four angles, and four vertices.

Question

How is a parallelogram defined based on its sides?

Answer

A parallelogram is a quadrilateral in which both pairs of opposite sides are parallel.

Question

According to Theorem 8.1, what is the result of drawing a diagonal in a parallelogram?

Answer

The diagonal divides the parallelogram into two congruent triangles.

Question

In the proof of Theorem 8.1, which side is common to triangles $\Delta ABC$ and $\Delta CDA$?

Answer

The diagonal $AC$ (or $CA$) is common to both triangles.

Question

Which congruence rule is utilized to prove that a diagonal divides a parallelogram into two congruent triangles?

Answer

The proof utilizes the $ASA$ (Angle-Side-Angle) congruence rule.

Question

What does Theorem 8.2 state regarding the lengths of sides in a parallelogram?

Answer

In a parallelogram, each pair of opposite sides is equal.

Question

State the converse of the property that opposite sides of a parallelogram are equal (Theorem 8.3).

Answer

If each pair of opposite sides of a quadrilateral is equal, then it is a parallelogram.

Question

What is the relationship between opposite angles in a parallelogram according to Theorem 8.4?

Answer

Each pair of opposite angles in a parallelogram is equal.

Question

Under what condition regarding angles is a quadrilateral considered a parallelogram by Theorem 8.5?

Answer

A quadrilateral is a parallelogram if each pair of opposite angles is equal.

Question

How do the diagonals of a parallelogram interact with one another according to Theorem 8.6?

Answer

The diagonals of a parallelogram bisect each other.

Question

If the diagonals of a quadrilateral bisect each other, what conclusion can be drawn about the shape?

Answer

The quadrilateral is a parallelogram.

Question

In the proof for Theorem 8.7, which congruence rule is used to show $\Delta AOB \cong \Delta COD$?

Answer

The $SAS$ congruence rule is implied (using bisected diagonals and vertically opposite angles).

Question

A rectangle is defined as a parallelogram that possesses at least one _____.

Answer

right angle

Question

If one angle of a rectangle is $90^\circ$, what is the measure of each of its other angles?

Answer

Each of the other angles is also $90^\circ$.

Question

Why do adjacent angles in a rectangle add up to $180^\circ$?

Answer

They are interior angles on the same side of a transversal between parallel lines.

Question

What is the geometric relationship between the diagonals of a rhombus?

Answer

The diagonals of a rhombus are perpendicular to each other.

Question

Which congruence rule is used to prove that the diagonals of a rhombus are perpendicular?

Answer

The $SSS$ congruence rule is used.

Question

In an isosceles triangle $ABC$ where $AB = AC$, what is the relationship between the exterior angle $\angle PAC$ and the interior angle $\angle ACB$?

Answer

The exterior angle $\angle PAC$ is equal to $2\angle ACB$.

Question

What specific shape is formed by the bisectors of the interior angles of two parallel lines intersected by a transversal?

Answer

The bisectors form a rectangle.

Question

What shape results from the intersection of the angle bisectors of a parallelogram?

Answer

The intersection of the angle bisectors forms a rectangle.

Question

In $\Delta ASD$, if $DS$ and $AS$ are angle bisectors of a parallelogram, what is the sum of $\angle DAS$ and $\angle ADS$?

Answer

The sum of the angles is $90^\circ$.

Question

If the diagonals of a parallelogram are equal in length, what type of quadrilateral is it?

Answer

It is a rectangle.

Question

What are the two primary diagonal properties that define a square?

Answer

The diagonals of a square are equal in length and bisect each other at right angles.

Question

If a diagonal of a parallelogram bisects one of its angles, what must be true about the quadrilateral's sides?

Answer

The quadrilateral is a rhombus, meaning all its sides are equal.

Question

State the Mid-point Theorem regarding the orientation of the joining segment.

Answer

The line segment joining the mid-points of two sides of a triangle is parallel to the third side.

Question

According to the Mid-point Theorem, how does the length of the segment joining two mid-points relate to the third side?

Answer

The segment is half the length of the third side.

Question

What does Theorem 8.9 state regarding a line through a mid-point parallel to another side?

Answer

A line drawn through the mid-point of one side of a triangle, parallel to another side, bisects the third side.

Question

By joining the mid-points of all three sides of a triangle, how many congruent triangles are created?

Answer

Four congruent triangles are created.

Question

If three parallel lines cut off equal intercepts on one transversal, what can be said about their intercepts on any other transversal?

Answer

They will cut off equal intercepts on any other transversal as well.

Question

In a quadrilateral $ABCD$, if $P, Q, R, S$ are the mid-points of the sides, what shape is $PQRS$?

Answer

The quadrilateral $PQRS$ is a parallelogram.

Question

The quadrilateral formed by joining the mid-points of the sides of a rhombus is a _____.

Answer

rectangle

Question

The quadrilateral formed by joining the mid-points of the sides of a rectangle is a _____.

Answer

rhombus

Question

In a trapezium where $AB \parallel DC$ and $E$ is the mid-point of $AD$, a line through $E$ parallel to $AB$ will intersect $BC$ at its _____.

Answer

mid-point

Question

In a parallelogram $ABCD$, if $E$ and $F$ are mid-points of $AB$ and $CD$, what do the segments $AF$ and $EC$ do to the diagonal $BD$?

Answer

The segments $AF$ and $EC$ trisect the diagonal $BD$.

Question

In a right-angled triangle, the line joining the mid-point of the hypotenuse to the right-angle vertex is _____ the length of the hypotenuse.

Answer

half

Question

Which property of diagonals is unique to both the rhombus and the square?

Answer

In both shapes, the diagonals bisect each other at right angles.

Question

Which property of diagonals is shared by both the rectangle and the square?

Answer

In both shapes, the diagonals are equal in length.

Question

What is the result of bisecting the exterior angle of an isosceles triangle and showing it is parallel to the base?

Answer

It helps prove that the resulting quadrilateral is a parallelogram.

Question

If line segment $DE = \frac{1}{2} AC$ and $DE \parallel AC$, what are $D$ and $E$ in relation to $\Delta ABC$?

Answer

They are the mid-points of two sides of the triangle.

Question

In Example 7, which theorem is applied twice to prove that $DE = EF$?

Answer

Theorem 8.9 (The converse of the Mid-point Theorem) is applied twice.

Question

What is the sum of the interior angles on the same side of a transversal intersecting two parallel lines?

Answer

The sum of the angles is $180^\circ$.

Question

If a quadrilateral is a parallelogram, what can be said about the areas of the two triangles formed by its diagonal?

Answer

The triangles are congruent, meaning they have equal areas.

Question

How is the converse of Theorem 8.6 (Theorem 8.7) stated?

Answer

If the diagonals of a quadrilateral bisect each other, then it is a parallelogram.

Question

To show a parallelogram is a rectangle, what is the minimum requirement regarding its angles?

Answer

It is sufficient to show that one of its angles is a right angle.

Question

In a rhombus, why is $\Delta AOD \cong \Delta COD$ by SSS?

Answer

Because $OA=OC$, $OD=OD$, and $AD=CD$ (all sides of a rhombus are equal).

Question

In Example 5, why is $\angle PSR = 90^\circ$?

Answer

It is vertically opposite to $\angle DSA$, which is $90^\circ$ due to the triangle angle sum property.

Question

A quadrilateral in which only one pair of opposite sides is parallel is called a _____.

Answer

trapezium

Question

If diagonal $AC$ of parallelogram $ABCD$ bisects $\angle A$, it also bisects _____.

Answer

$\angle C$

Question

The segments $SR$ and $PQ$ in a mid-point quadrilateral are both _____ to diagonal $AC$.

Answer

parallel

Question

In the Mid-point Theorem, $EF$ is parallel to $BC$ and $EF = \frac{1}{2}$ _____.

Answer

$BC$

Question

What is the first step in proving that the bisectors of the angles of a parallelogram form a rectangle?

Answer

Showing that the sum of half of two adjacent interior angles is $90^\circ$.

Question

In the proof for Theorem 8.8, what shape is $BCDE$ constructed to be?

Answer

It is constructed to be a parallelogram.

Question

According to the summary, the line-segment joining the mid-points of any two sides of a triangle is half of the _____.

Answer

third side

Question

In a rectangle, the diagonals bisect each other and are _____.

Answer

equal

Question

What property makes a parallelogram a rhombus?

Answer

A parallelogram is a rhombus if all four of its sides are equal.

Question

When using ASA to prove Theorem 8.1, what type of angles are $\angle BCA$ and $\angle DAC$?

Answer

They are a pair of alternate interior angles.