# 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 t****he ****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.

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