In the kite, show that the diagonal (i) bisects and , (ii) bisects the diagonal , that is, , and is perpendicular to it. Hint: Is ?

We will use triangle congruence to prove the properties of the kite's diagonals. Step 1 — Bisecting angles ADC and ABC Let us look at the quadrilateral ABCD. The tick marks show that adjacent sides are equal. So, AD = CD and AB = CB. Consider the triangles and . We know that . We also know that . The side is common to both triangles. By the SSS (Side-Side-Side) congruence criterion, . Since the triangles are congruent, their corresponding angles are equal. This means . So, bisects . It also means . So, bisects . Step 2 — Bisecting diagonal AC and being perpendicular to it Now, let us consider the triangles and . From Step 1, we know that bisects . This means . By the SAS (Side-Angle-Side) congruence criterion, . Since the triangles are congruent, their corresponding parts are equal. This means . So, bisects the diagonal . It also means . Angles and form a linear pair. This means their sum is . Since , we can write: This means is perpendicular to . Answer > (i) We showed that using the SSS congruence criterion. From this congruence, we concluded that and . Therefore, the diagonal bisects and . > (ii) We showed that using the SAS congruence criterion. From this congruence, we concluded that , meaning bisects the diagonal . We also found that . Since these angles form a linear pair, each must be , which means is perpendicular to .

Quadrilaterals | IT

Question 42

In the kite, show that the diagonal $BD$

(i) bisects $\angle ABC$ and $\angle ADC$ ,

(ii) bisects the diagonal $AC$ , that is, $AO = OC$ , and is perpendicular to it.

Hint: Is $\Delta AOB \cong \Delta COB$ ?

Solution

We will use triangle congruence to prove the properties of the kite's diagonals.

Step 1 — Bisecting angles ADC and ABC

Let us look at the quadrilateral ABCD. The tick marks show that adjacent sides are equal. So, AD = CD and AB = CB.

Consider the triangles $\triangle ABD$ and $\triangle CBD$ . We know that $AD = CD$ . We also know that $AB = CB$ . The side $BD$ is common to both triangles.

$AD = CD \quad (\text{Given})$

$AB = CB \quad (\text{Given})$

$BD = BD \quad (\text{Common side})$

By the SSS (Side-Side-Side) congruence criterion, $\triangle ABD \cong \triangle CBD$ . Since the triangles are congruent, their corresponding angles are equal. This means $\angle ADB = \angle CDB$ . So, $BD$ bisects $\angle ADC$ . It also means $\angle ABD = \angle CBD$ . So, $BD$ bisects $\angle ABC$ .

Diagram 1

Step 2 — Bisecting diagonal AC and being perpendicular to it

Now, let us consider the triangles $\triangle AOB$ and $\triangle COB$ . From Step 1, we know that $BD$ bisects $\angle ABC$ . This means $\angle ABO = \angle CBO$ .

$AB = CB \quad (\text{Given})$

$\angle ABO = \angle CBO \quad (\text{From Step 1, BD bisects } \angle ABC)$

$BO = BO \quad (\text{Common side})$

By the SAS (Side-Angle-Side) congruence criterion, $\triangle AOB \cong \triangle COB$ . Since the triangles are congruent, their corresponding parts are equal. This means $AO = OC$ . So, $BD$ bisects the diagonal $AC$ . It also means $\angle AOB = \angle COB$ .

Angles $\angle AOB$ and $\angle COB$ form a linear pair. This means their sum is $180^\circ$ . $\angle AOB + \angle COB = 180^\circ$ Since $\angle AOB = \angle COB$ , we can write: $2 \times \angle AOB = 180^\circ$ $\angle AOB = \frac{180^\circ}{2}$ $\angle AOB = 90^\circ$ This means $BD$ is perpendicular to $AC$ .

Diagram 2

Answer

(i) We showed that $\triangle ABD \cong \triangle CBD$ using the SSS congruence criterion. From this congruence, we concluded that $\angle ADB = \angle CDB$ and $\angle ABD = \angle CBD$ . Therefore, the diagonal $BD$ bisects $\angle ABC$ and $\angle ADC$ . (ii) We showed that $\triangle AOB \cong \triangle COB$ using the SAS congruence criterion. From this congruence, we concluded that $AO = OC$ , meaning $BD$ bisects the diagonal $AC$ . We also found that $\angle AOB = \angle COB$ . Since these angles form a linear pair, each must be $90^\circ$ , which means $BD$ is perpendicular to $AC$ .

More questions in IT

Observe the following figures.

Figs. (i), (ii), and (iii) are quadrilaterals, and the others are not. Why?

Are there other ways to define a rectangle?

A Carpenter's Problem

A carpenter needs to put together two thin strips of wood, as shown in Fig. 1, so that when a thread is passed through their endpoints, it forms a rectangle. She already has one 8 cm long strip. What should be the length of the other strip? Where should they both be joined?

Let us first model the structure that the carpenter has to make. The strips can be modelled as line segments. They are the diagonals of the quadrilateral formed by their endpoints. For the quadrilateral to be a rectangle, we need to answer the following questions —

What is the length of the other diagonal?
What is the point of intersection of the two diagonals?
What should the angle be between the diagonals?

Can the following equalities be used to establish that $\Delta AOD \cong \Delta COB$ ?

$AO = CO$ (proved above)
$\angle AOB = \angle COD$ (vertically opposite angles)
$AD = CB$

Context: Let us check what quadrilateral we get if we draw the two diagonals such that their lengths are equal, they bisect each other and have an arbitrary angle, say $60^\circ$ , between them as shown in the figure to the right.

Q. Can you find all the remaining angles?

Context: In $\Delta AOB$ , since $OA = OB$ , the angles opposite them are equal, say $a$ .

Q. Can you find the value of $a$ ?

Can we now identify what type of quadrilateral ABCD is?

Notice that its angles all add up to 90° (30° + 60°).

What can we say about its sides?

Will ABCD remain a rectangle if the angles between the diagonals are changed? Can we generalise this?

Take one of the angles between the diagonals as $x$ .

Q10

Context: We can compute the four angles between the diagonals to be $x, x, 180 - x,$ and $180 - x.$

Q. Can you find the other angles?

Q11

Context: Since we know that $\Delta AOB$ is isosceles, we can denote the measures of both of its base angles by $a$ .

Q. What is the value of $a$ (in degrees) in terms of $x$ ?

Q12

Context: Thus, all four angles of the quadrilateral ABCD are 90°.

Q. What can we say about AB and CD, and AD and BC?

Q13

In the earlier definition, we stated that a rectangle has (a) opposite sides of equal length, and (b) all angles equal to 90°. Would we be wrong if we just define a rectangle as a quadrilateral in which all the angles are 90°?

Q14

If you think that this definition is incomplete, try constructing a quadrilateral in which the angles are all 90° but the opposite sides are not equal.

Are you able to construct such a quadrilateral?

Q15

Is it wrong to write ΔBAD ≅ ΔCDB? Why?

Q16

Can you similarly show that AB is parallel to DC (AB || DC)?

Q17

In the quadrilaterals below, are there any non-rectangles?

Q18

Let us consider the Carpenter's Problem again. If the wooden strips have to be placed such that the thread passing through their endpoints forms a square, what must be done?

Q19

What more needs to be done to get equal sidelengths as well? Can this be achieved by properly choosing the angle between the diagonals? See if you can reason and/or experiment to figure this out!

Q20

Can this be used to find the angles $\angle\text{BOA}$ and $\angle\text{BOC}$ formed by the diagonals?

Q21

Context: The diagonals of a square are of equal lengths and bisect each other at right angles.

Q. Using this fact, construct a square with a diagonal of length 8 cm.

Q22

Context: Since a square is a special type of rectangle, all the properties of a rectangle hold true for a square.

Q. Verify if this is true by going through geometric reasoning in Deduction 1 and Deduction 2, and see if they apply to a square as well.

Q23

Q. Similarly, find $\angle 2$ and $\angle 4$ .

Q24

4.2 Angles in a Quadrilateral

Is it possible to construct a quadrilateral with three angles equal to 90° and the fourth angle not equal to 90°?

Q25

But why not?

Q26

Are there quadrilaterals that have parallel opposite sides that are not rectangles?

Q27

Construct such a figure by recalling how parallel lines can be constructed using a ruler and a set-square, or a compass and a ruler.

Q28

Context: Consider a parallelogram $ABCD$ with adjacent sides of lengths $4\text{ cm}$ and $5\text{ cm}$ , and an angle of $30^\circ$ between them.

Q. What are the remaining angles of the parallelogram? What are the lengths of the remaining sides? See if you can reason out and/or experiment to figure these out.

Q29

Deduction 7— What can we say about the sides of a parallelogram?

By looking at a parallelogram, it appears that the opposite sides are equal. Can we again use congruence to show this? Which two triangles can be considered for this?

Q30

Is it wrong to write $\Delta\text{ABD} \cong \Delta\text{CBD}$ ? Why?