RD Sharma 2020 solution class 9 chapter 12 Congruent Triangles Exercise 12.5

Exercise 12.5

Page-12.61

Question 1:

ABC is a triangle and D is the mid-point of BC. The perpendiculars from D to AB and AC are equal. Prove that the triangle is-isosceles.

Answer 1:

We have to prove thatis isosceles.

Let and be perpendicular from D on AB and AC respectively.

In order to prove that

We will prove that

Now in and we have

(Since D is mid point of BC)

(Given)

So by congruence criterion we have 

And

Hence is isosceles.

Question 2:

ABC is a triangle in which BE and CF are, respectively, the perpendiculars to the sides AC and AB. If BE = CF, prove that ΔABC is isosceles.

Answer 2:

It is given that 

, and

And.

 

We have to prove is isosceles.

To prove is isosceles we will prove

For this we have to prove

Now comparing and we have

(Given)

(Common side)

So, by right hand side congruence criterion we have 

So (since sides opposite to equal angle are equal)

Hence is isosceles.

Question 3:

If perpendiculars from any point within an angle on its arms are congruent, prove that it lies on the bisector of that angle.

Answer 3:

Let P be a point within such that

 

 

We have to prove that P lies on the bisector of

In and we have

(We have)

(Common)

So by right hand side congruence criterion, we have 

So,

Hence P lies on the bisector of proved.

Question 4:

In the given figure, AD ⊥ CD and CB ⊥ CD. If AQ = BP and DP = CQ, prove that
∠DAQ = ∠CBP.
 

Answer 4:

It is given that 

, and

If and

We have to prove that

In triangles and we have

(Since given)

So

And (given)

So by right hand side congruence criterion we have

So

HenceProved.

Page-12.62

Question 5:

Which of the following statements are true (T) and which are false (F):

(i) Sides opposite to equal angles of a triangle may be unequal.

(ii) Angles opposite to equal sides of a triangle are equal.

(iii) The measure of each angle of an equilateral triangle is 60°.

(iv) If the altitude from one vertex of a triangle bisects the opposite side, then the triangle may be isosceles.

(v) The bisectors of two equal angles of a triangle are equal.

(vi) If the bisector of the vertical angle of a triangle bisects the base, then the triangle may be isosceles.

(vii) The two altitudes corresponding to two equal sides of a triangle need not be equal.

(viii) If any two sides of a right triangle are respectively equal to two sides of other right triangle, then the two triangles are congruent.

(ix) Two right triangles are congruent if hypotenuse and a side of one triangle are respectively equal equal to the hypotenuse and a side of the other triangle.

 

 

 

Answer 5:

(1)

(2)

(3)

(4)

(5)

(6)

(7)

(8)

(9)

Question 6:

Fill in the blanks in the following so that each of the following statements is true.

(i) Sides opposite to equal angles of a triangle are .......

(ii) Angle opposite to equal sides of a triangle are .......

(iii) In an equilateral triangle all angles are ........

(iv) In a ABC if ∠A = ∠C, then AB =  ............

(v) If altitudes CE and BF of a triangle ABC are equal, then AB = ........

(vi) In an isosceles triangle ABC with AB = AC, if BD and CE are its altitudes, then BD is ...... CE.

(vii)  In right triangles ABC and DEF, if hypotenuse AB = EF and side AC = DE, then ΔABC ≅ Δ .........

Answer 6:

(1)

(2)

(3)

(4)

(5)

(6)

(7)

Question 7:

ABCD is a square, X and Y are points on sides AD and BC respectively such that AY= BX. Prove that BY = AX and ∠BAY = ∠ABX.

Answer 7:

It is given ABCD is a square and

We have to prove that and

In right angled trianglesand Δ we have

And, and 

So by right hand side congruence criterion we have 

So (since triangle is congruent)

HenceProved.

Question 8:

ABCD is a quadrilateral such that AB = AD  and CB = CD. Prove that AC is the perpendicular bisector of BD.

Answer 8:

In quadrilateral ABCD, AB = AD and BC = CD. Let AC and BD intersect at O.

In ∆ABC and ∆ADC,

AB = AD        (Given)

AC = AC        (Common)

BC = CD        (Given)

∴ ∆ABC ≅ ∆ADC       (SSS congruence criterion)

⇒ ∠BAC = ∠DAC       (CPCT)

Now, in ∆ABO and ∆ADO,

AB = AD                   (Given)

∠BAO = ∠DAO        (Proved above)

AO = AO                   (Common)

∴ ∆ABO ≅ ∆ADO       (SAS congruence axiom)

⇒ OB = OD             .....(1)      (CPCT)

∠AOB = ∠AOD       .....(2)      (CPCT)

Now,

∠AOB + ∠AOD = 180º         (Linear pair)

⇒ 2∠AOB = 180º          [Using (2)]

⇒ ∠AOB = 90º        .....(3)

From (1) and (3), we conclude that AC is the perpendicular bisector of BD.

Question 9:

∆ABC is a right triangle such that AB = AC and bisector of angle C intersects the side AB at D. Prove that AC + AD = BC.

Answer 9:

∆ABC is a right triangle such that AB = AC. CD is the bisector of ∠C which intersects AB at D.

Let AB = AC = x units

In right ∆ABC,

${\mathrm{BC}}^2={\mathrm{AB}}^2+{\mathrm{AC}}^2$        (Pythagoras theorem)

$\Rightarrow {\mathrm{BC}}^2=x^2+x^2=2x^2$

$\Rightarrow \mathrm{BC}=\sqrt{2}x$

In ∆ABC, CD is the bisector of ∠C.

$\therefore \frac{\mathrm{AD}}{\mathrm{BD}}=\frac{\mathrm{AC}}{\mathrm{BC}}$          (The bisector of an angle of a triangle divides the opposite side in the ratio of sides containing the angle)

$\Rightarrow \frac{\mathrm{AD}}{x-\mathrm{AD}}=\frac{x}{\sqrt{2}x}$

$\Rightarrow \sqrt{2}\mathrm{AD}=x-\mathrm{AD}$

$\Rightarrow \mathrm{AD}\left(\sqrt{2}+1\right)=x$

$\Rightarrow \mathrm{AD}=\frac{x}{\sqrt{2}+1}=x\left(\sqrt{2}-1\right)$

Now,

$\mathrm{AC}+\mathrm{AD}=x+\sqrt{2}\left(x-1\right)=\sqrt{2}x=\mathrm{BC}$

Hence proved.

Question 10:

O is a point in the interior of a square ABCD such that OAB is an equilateral triangle. Show that ∆OCD is an isosceles triangle.

Answer 10:

O is a point in the interior of square ABCD. ∆OAB is an equilateral triangle.

Now,

∠DAB = ∠CBA        .....(1)         (Measure of each angle of a square is 90º)

∠OAB = ∠OBA        .....(2)         (Measure of each angle of an equilateral triangle is 60º)

Subtracting (2) from (1), we get

∠DAB − ∠OAB = ∠CBA − ∠OBA

⇒ ∠OAD = ∠OBC  

In ∆OAD and ∆OBC,

OA = OB      (Sides of an equilateral triangle are equal)

∠OAD = ∠OBC          (Proved above)

AD = BC       (Sides of a square are equal)

∴ ∆OAD ≅ ∆OBC       (SAS congruence axiom)

⇒ OD = OC             (CPCT)

In ∆OCD,

OC = OD

∴ ∆OCD is an isosceles triangle.    (A triangle whose two sides are equal is an isosceles triangle)

Question 11:

ABCD is a quadrilateral in which AB = BC and AD = CD. Show that BD bisects both the angles ∠ABC and ∠ADC.

Answer 11:

In quadrilateral ABCD, AB = BC and AD = CD.

In ∆ABD and ∆CBD,

AB = CB        (Given)

BD = BD        (Common)

AD = CD        (Given)

∴ ∆ABD ≅ ∆CBD       (SSS congruence criterion)

So, ∠ABD = ∠CBD       .....(1)       (CPCT)

∠ADB = ∠CDB             .....(2)       (CPCT)

From (1) and (2), we conclude that

BD bisects both ∠ABC and ∠ADC.

Question 12:

Line segment joining the mid-points M and N of parallel sides AB and DC, respectively of a trapezium ABCD is perpendicular to both the sides AB and DC. Prove that AD = BC.

Answer 12:

ABCD is a trapezium with AB || CD. M and N are the mid-points of sides AB and AC, respectively.

Join AN and BN.

In ∆AMN and ∆BMN,

AM = BM                      (M is the mid-point of AB)

∠AMN = ∠BMN           (MN ⊥ AB)

MN = MN                      (Common)

∴ ∆AMN ≅ ∆BMN        (SAS congruence axiom)

So, AN = BN         .....(1)       (CPCT)

∠ANM = ∠BNM            (CPCT)

Now,

∠DNM = ∠CNM            (90º each)

∴ ∠DNM − ∠ANM = ∠CNM − ∠BNM

⇒ ∠AND = ∠BNC      .....(2)

In ∆AND and ∆BNC,

DN = CN                       (N is the mid-point of CD)

∠AND = ∠BNC           [From (2)]

AN = BN                      [From (1)]

∴ ∆AND ≅ ∆BNC        (SAS congruence axiom)

So, AD = BC                  (CPCT)

Hence proved.