Class 9 RD Sharma Solution – Chapter 10 Congruent Triangles- Exercise 10.5

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.

Solution:

Given: D is the mid-point of BC so, BD = DC, ED = FD, and ED ⊥ AB, FD ⊥ AC, so ED = FD 

Prove: ΔABC is an isosceles triangle

In ΔBDE and ΔCDF

ED = FD [Given]

BD = DC [D is mid-point]

∠BED = ∠CFD = 90°

By RHS congruence criterion

ΔBDE ≅ ΔCDF

So, now by C.P.C.T

BE = CF … (i) 

Now, in △AED and △AFD

ED = FD [Given]

AD = AD [Common]

∠AED = ∠AFD = 90°

By RHS congruence criterion

△AED ≅ △AFD

So, now by C.P.C.T

So, EA = FA … (ii) 

Now by adding equation (i) and (ii), we get

BE + EA = CF + FA

AB = AC

So, ΔABC is an isosceles triangle because two sides of the triangle are equal.

Hence proved

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.

Solution:

Given: BE ⊥ AC, CF ⊥ AS, BE = CF. 

To prove: ΔABC is isosceles

In ΔBCF and ΔCBE,

∠BFC = CEB = 90° [Given]

BC = CB [Common side]

And CF = BE [Given]

By RHS congruence criterion

ΔBFC ≅ ΔCEB  

So, now by C.P.C.T

∠FBC = ∠EBC   

∠ABC = ∠ACB 

and AC = AB [Because opposite sides to equal angles are equal]

So, ΔABC is isosceles

Hence proved

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.

Solution:

Let us consider ∠ABC and BP is an arm within∠ABC

So now draw perpendicular from point P on arm BA and BC, i.e., PN and PM

Prove: BP is the angular bisector of ∠ABC.

In ΔBPM and ΔBPN

∠BMP = ∠BNP = 90° [Given]

MP = NP [Given]

BP = BP [Common side]

So, by RHS congruence criterion

ΔBPM ≅ ΔBPN 

So, by C.P.C.T

∠MBP = ∠NBP 

and BP is the angular bisector of ∠ABC.

Hence proved

Question 4. In figure, AD ⊥ CD and CB ⊥ CD. If AQ = BP and DP = CQ, prove that ∠DAQ = ∠CBP.

Solution:

Given that AD ⊥ CD, CB ⊥ CD, AQ = BP and DP = CQ,

Prove:∠DAQ = ∠CBP

We have DP = CQ

So by adding PQ on both sides, we get

DP + PQ = CQ + PQ

DQ = CP … (i)

In ΔDAQ and ΔCBP

We have

∠ADQ = ∠BCP = 90° [Given]

And DQ = PC [From (i)]

So, by RHS congruence criterion

ΔDAQ ≅ ΔCBP 

So, by C.P.C.T

∠DAQ = ∠CBP 

Hence proved

Question 5. 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.

Solution:

In ABCD square, 

X and Y are points on sides AD and BC 

So, AY = BX.

To prove: BY = AX and ∠BAY = ∠ABX

Now, join Band X, A and Y

So, 

∠DAB = ∠CBA = 90° [Given ABCD is a square]

Also, ∠XAB = ∠YAB = 90° 

In ΔXAB and ΔYBA

∠XAB = ∠YBA = 90° [given]

AB = BA [Common side]

So, by RHS congruence criterion

ΔXAB ≅ ΔYBA  

So, by C.P.C.T

BY = AX 

∠BAY = ∠ABX 

Hence proved

Question 6. 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-angled triangles are congruent if the hypotenuse and a side of one triangle are respectively equal to the hypotenuse and a side of the other triangle.

Solution:

(i) False 

(ii) True

(iii) True 

(iv) False 

(v) True 

(vi) False

(vii) False 

(viii) False

(ix) True 

Question 7. Fill 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 ΔABC, 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 ___

Solution:

(i) Equal

(ii) Equal

(iii) Equal 

(iv) AB = BC

(v) AB = AC

(vi) BD is equal to CE

(vii) ΔABC ≅ ΔEFD.