RD Sharma 2020 solution class 9 chapter 12 Congruent Triangles Exercise 12.1

Exercise 12.1

Page-12.15

Question 1:

In the given figure, the sides BA  and CA have been produced such that BA = AD and CA = AE. Prove that segment DE || BC.

Answer 1:

It is given that

We have to prove that

Now considering the two triangles we have

In

(Given)

(Given)

We need to show to prove.

Now

(Vertically opposite angle)

So by congruence criterion we have

So and 

Then

, and

Hence from above conditions .

Question 2:

In a ΔPQR, if PQ =  QR and L, M and N are the mid-points of the sides PQ, OR, and RP respectively. Prove that LN = MN.
 

Answer 2:

It is given that

And is the mid point of

So

And is the mid point of

So

And is the mid point of

So

We have to prove that

In we have 

(Equilateral triangle)

Then 

, and

, and

Similarly comparing and we have

, and

And (Since N is the mid point of )

So by congruence criterion, we have 

Hence.

Question 3:

Prove that the medians of an equilateral triangle are equal.

Answer 3:

We have to prove that the median of an equilateral triangle are equal.

Let be an equilateral triangle with as its medians.

Let

In we have

(Since similarly)

(In equilateral triangle, each angle)

And (common side)

So by congruence criterion we have

This implies that, 

Similarly we have

Hence .

Question 4:

In a Δ ABC, if ∠A = 120° and AB = AC. Find ∠B and ∠C.

Answer 4:

In, it is given that

, and

We have to find, and

Since and

Then (as AB = AC)

Now 

(By property of triangle)

Thus, 

, as (given)

So,

Since,, so

Hence .

Question 5:

In a ΔABC, if AB = AC and ∠B = 70°, find ∠A.

Answer 5:

In it is given that

, and

We have to find.

Since

Then (isosceles triangles)

Now 

(As given)

Thus 

(Property of triangle)

Hence .

Question 6:

The vertical angle of an isosceles triangle is 100°. Find its base angles.

Answer 6:

Suppose in the isosceles triangle ΔABC it is given that

We have to find the base angle.

Now vertical angle (given)

And

Since then

Now

(By property of triangle)

So 

Hence the base angle is.

Question 7:

In the given figure, AB = AC and ∠ACD = 105°, find ∠BAC.
 

Answer 7:

It is given that

We have to find.

(Isosceles triangle)

Now 

Since exterior angle of isosceles triangle is the sum of two internal base angles

Now 

So, (By property of triangle)

Hence .

Question 8:

Find the measure of each exterior angle of an equilateral triangle.

Answer 8:

We have to find the measure of each exterior angle of an equilateral triangle.

It is given that the triangle is equilateral

So, and 

Since triangle is equilateral 

So,

Now we have to find the exterior angle.

As we know that exterior angle of the triangle is sum of two interior angles

Thus

Hence each exterior angle is.

Question 9:

If the base of an isosceles triangle is produced on both sides, prove that the exterior angles so formed are equal to each other.

Answer 9:

It is given that the base of an isosceles triangle is produced on both sides.

We have to prove that the exterior angles so formed are equal to each other.

That is we need to show that

Let the is isosceles having base and equal sides AB and AC

Then, and

(Isosceles triangles)

Now 

    .........(1)

And, 

   .......(2)

Thus 

  ........(3)

Now from equation (2)

     .........(4)

Since

Hence from equation (3) and (4)

Question 10:

In the given figure, AB = AC and DB =  DC, find the ratio ∠ABD : ∠ACD.

Answer 10:

It is given that

We have to find the ratio.

Since

And

So we have,

So

Hence .

Page-12.16

Question 11:

Determine the measure of each of the equal angles of a right-angled isosceles triangle.

OR

ABC is a right-angled triangle in which ∠A = 90° and AB = AC. Find ∠B and ∠C.
 

Answer 11:

It is given that

Is right angled triangle

And 

We have to find and

Since

(Isosceles triangle)

Now 

(Property of triangle)

()

So

Hence 

Question 12:

In the given figure, PQRS is a square and SRT is an equilateral triangle. Prove that
(i) PT = QT
(ii) ∠TQR = 15°

Answer 12:

It is given that

Δis a square and Δ is an equilateral triangle.

We have to prove that

(1) and (2)

(1)

Since,

(Angle of square)

(Angle of equilateral triangle)

Now, adding both

Similarly, we have

Thus in and we have

(Side of square)

And (equilateral triangle side)

So by congruence criterion we have

Hence.

(2)
Since 
QR = RS ( Sides of Square)
RS = RT (Sides of Equilateral triangle)

We get
QR = RT

Thus, we get
$\angle TQR=\angle RTQ$  (Angles opposite to equal sides are equal)

Now, in the triangle TQR, we have

$$\begin{gathered} \angle TQR+\angle RTQ+\angle QRT={180}^0 \\ \angle TQR+\angle TQR+{150}^0={180}^0 \\ 2\angle TQR+{150}^0={180}^0 \\ 2\angle TQR={180}^0-{150}^0 \\ 2\angle TQR={30}^0 \\ \angle TQR=\frac{{30}^0}{2}={15}^0 \end{gathered}$$

Question 13:

AB is a line segment. P and Q are points on opposite sides of AB such that each of them is equidistant from the points A and B (See Figure). Show that the line PQ is perpendicular bisector of AB.
 

Answer 13:

It is given that

P and Q are equidistant from A and B that is

, and

We are asked to show that line PO is perpendicular bisector of line AB.

First of all we will show that ΔAQP and ΔQBP are congruent to each other and ultimately we get the result.

Consider the triangles AQP and QBP in which

AP=BP, AQ=BQ, PQ=PQ

So by SSS property we have

Implies that

Now consider the triangles ΔAPC and ΔPCB in which

And

So by SAS criterion we find that,

So this implies that AC=BC and

But 

Hence PQ is perpendicular bisector of AB.

Question 14:

In a ∆ABC, D is the mid-point of AC such that BD = $\frac{1}{2}$ AC. Show that ∠ABC is a right angle.

Answer 14:

In a ∆ABC, D is the mid-point of AC such that BD = $\frac{1}{2}$AC.

D is the mid-point of AC.

∴ AD = CD = $\frac{1}{2}$AC 

⇒ AD = CD = BD           (BD = $\frac{1}{2}$AC)

In ∆ABD,

AD = BD

∴ ∠ABD = ∠A          .....(1)      (In a triangle, equal sides have equal angles opposite to them)

In ∆CBD,

CD = BD

∴ ∠CBD = ∠C           .....(2)     (In a triangle, equal sides have equal angles opposite to them)

Adding (1) and (2), we get

∠ABD + ∠CBD = ∠A + ∠C

⇒ ∠B = ∠A + ∠C     .....(3)

In ∆ABC,

∠A + ∠B + ∠C = 180º        (Angle sum property of triangle)

⇒ ∠B + ∠B = 180º             [Using (3)]

⇒ 2∠B = 180º 

⇒ ∠B = $\frac{180°}{2}$ = 90º

Thus, ∠ABC is a right angle.

Question 15:

∆ABC is a right triangle with AB = AC. Bisector of ∠A meets BC at D. Prove that BC = 2AD.

Answer 15:

In ∆ABC,

AB = AC     (Given)

∴ ∠C = ∠B      .....(1)       (In a triangle, angles opposite to equal sides are equal)

Also, ∠A = 90º     (Given)

Now,

∠A + ∠B + ∠C = 180º    (Angle sum property of triangle)

⇒ 90º + 2∠B = 180º        [Using (1)]

⇒ 2∠B = 180º − 90º = 90º

⇒ ∠B = $\frac{90°}{2}$ = 45º

∴ ∠C = ∠B = 45º   

It is given that, AD is the bisector of ∠A.

∴ ∠CAD = ∠BAD$=\frac{\angle \mathrm{A}}{2}=\frac{90°}{2}$ = 45º

In ∆ABD,

∠B = ∠BAD      (Each measure 45º)

∴ AD = BD       .....(2)     (In a triangle, sides opposite to equal angles are equal)
 
In ∆ACD,

∠C = ∠CAD      (Each measure 45º)

∴ AD = CD       .....(3)     (In a triangle, sides opposite to equal angles are equal)

Adding (2) and (3), we have

AD + AD = BD + CD

⇒ 2AD = BC

Or BC = 2AD

Hence proved.

Question 16:

∆ABC is a right triangle right angled at B such that ∠BCA = 2∠BAC. Show that AC = 2BC.

Answer 16:

∆ABC is a right triangle right angled at B such that ∠BCA = 2∠BAC. 

Produce CB to D such that BC = BD.

In ∆ABD and ∆ABC,

BD = BC                   (Construction)

∠ABD = ∠ABC       (90º each)

AB = AB                   (Common)

∴∆ABD ≅ ∆ABC     (SAS congruence axiom)

So, AD = AC             .....(1)         (CPCT)

∠BAD = ∠BAC       .....(2)          (CPCT)

Now,

∠BCA = 2∠BAC

⇒ ∠BCA = ∠BAC + ∠BAC

⇒ ∠BCA = ∠BAC + ∠BAD      [Using (2)]

⇒ ∠BCA = ∠CAD

In ∆ACD,

∠DCA = ∠CAD         (Proved above)

⇒ AD = CD                (Sides opposite to equal angles in a triangle are equal)

⇒ AC = BC + BD      [Using (1)]

⇒ AC = BC + BC      (BC = BD)

⇒ AC = 2BC