Isosceles Triangles [Including Inequalities] — Test Yourself

∠ABC = 90° and P is a point on side AC. Then:

1. PA = PB

2. PA > PB

3. PA < PB

4. none of these

Answer

In the right angled triangle ABC,

Let, ∠ACB = x

From figure,

⇒ ∠PCB = ∠ACB = x

⇒ ∠PBC = ∠PCB = x

By angle sum property of triangle,

⇒ ∠ABC + ∠ACB + ∠BAC = 180°

⇒ 90° + x + ∠BAC = 180°

⇒ ∠BAC = 180° - 90° - x = 90° - x

⇒ ∠BAP = ∠BAC = 90° - x.

From figure,

⇒ ∠ABP = ∠ABC - ∠PBC = 90° - x.

∴ ∠BAP = ∠ABP = 90° - x ...................(1)

In △ BAP,

⇒ ∠BAP = ∠ABP [From (1)]

∴ PB = PA (Sides opposite to equal angles in a triangle are equal)

Hence, option 1 is the correct option.

Triangles ABC is equilateral and BC = CE, then angle AEC is:

1. 60°

2. 45°

3. 30°

4. 120°

Answer

Since, ABC is an equilateral triangle, ∠A = ∠B = ∠C = 60°.

From figure,

⇒ ∠ACB + ∠ACE = 180° [Linear pairs]

⇒ 60° + ∠ACE = 180°

⇒ ∠ACE = 120°.

From figure, BC = CE (Given) ....................(1)

⇒ BC = AC (Side of equilateral triangle) ...................(2)

⇒ AC = CE (From equation (1) and (2))

⇒ ∠AEC = ∠CAE = y (let) [As angles opposite to equal sides of a triangle are equal]

By angle sum property in triangle AEC,

⇒ ∠AEC + ∠CAE + ∠ACE = 180°

⇒ y + y + 120° = 180°

⇒ 2y = 60°

⇒ y = 30°.

∴ ∠AEC = 30°.

Hence, option 3 is the correct option.

Side BA is produced upto point D and side BC upto point E such that ∠DAC = 110° and ∠ACE = 125°. Then the largest side of the triangle ABC is

1. AB

2. BC

3. AC

4. none of these

Answer

From figure, ∠DAC and ∠BAC forms linear pair.

⇒ ∠DAC + ∠BAC = 180°

⇒ 110° + ∠BAC = 180°

⇒ ∠BAC = 180° - 110°

⇒ ∠BAC = 70°

Similarly, ∠ACE and ∠ACB forms linear pair.

⇒ ∠ACE + ∠ACB = 180°

⇒ 125° + ∠ACB = 180°

⇒ ∠ACB = 180° - 125°

⇒ ∠ACB = 55°

In ΔABC, according to angle sum property,

⇒ ∠ABC + ∠ACB + ∠BAC = 180°

⇒ ∠ABC + 55° + 70° = 180°

⇒ ∠ABC + 125° = 180°

⇒ ∠ABC = 180° - 125°

⇒ ∠ABC = 55°

As we know that the side opposite to the largest angle in a triangle is the longest side.

Since, ∠BAC is the largest angle of the triangle.

So, BC is the largest side of the triangle ABC.

Hence, option 2 is the correct option.

In the given figure,

1. AC = CD

2. AB > CD

3. AB < CD

4. none of these

Answer

Since, AB = AC.

∴ ∠ABC = ∠ACB = 70° (As angles opposite to equal sides of an isosceles triangle are equal.)

From figure,

⇒ ∠ACB + ∠ACD = 180° [Linear pairs]

⇒ 70° + ∠ACD = 180°

⇒ ∠ACD = 110°.

In △ACD,

⇒ ∠CAD + ∠ADC + ∠ACD = 180°

⇒ ∠CAD + 40° + 110° = 180°

⇒ ∠CAD + 150° = 180°

⇒ ∠CAD = 30°.

In △ACD,

∠ADC = 40°

∠CAD = 30°

∴ ∠ADC > ∠CAD

∴ AC > CD (As side opposite to greater angle is greater.)

Since, AB = AC,

∴ AB > CD.

Hence, option 2 is the correct option.

Statement (1): In the given figure, AC = DC = BD and ∠B = 30°. Then ΔCAD will be equilateral.

Statement (2): ∠CDA = ∠CAD

= 30° + 30° = 60°

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

In ΔBDC,

⇒ DC = DB

⇒ ∠DBC = ∠DCB = 30° (Angles opposite to equal sides of a triangle are always equal)

In ΔBDC, according to angle sum property,

⇒ ∠BDC + ∠DBC + ∠DCB = 180°

⇒ ∠BDC + 30° + 30° = 180°

⇒ ∠BDC + 60° = 180°

⇒ ∠BDC = 180° - 60°

⇒ ∠BDC = 120°

Since ∠BDC and ∠CDA forms linear pair.

⇒ ∠BDC + ∠CDA = 180°

⇒ 120° + ∠CDA = 180°

⇒ ∠CDA = 180° - 120°

⇒ ∠CDA = 60°

Since, AC = DC

⇒ ∠CDA = ∠CAD = 60° (Angles opposite to equal sides of a triangle is always equal)

In ΔACD, according to angle sum property,

⇒ ∠ACD + ∠CDA + ∠CAD = 180°

⇒ ∠ACD + 60° + 60° = 180°

⇒ ∠ACD + 120° = 180°

⇒ ∠ACD = 180° - 120°

⇒ ∠ACD = 60°

So, ΔCAD is equilateral.

∴ Both the statements are true.

Hence, option 1 is the correct option.

Statement (1): AB = AC and D is any point in side BC of △ABC. Then AB will be greater than AD.

Statement (2): AB = AC

⇒ ∠C = ∠B

In ΔADC, ∠ADB = ∠C + ∠DAC

⇒ ∠ADB > ∠C

⇒ ∠ADB > ∠B

So, AB > AD

1. Both the statements are true.

2. Both the statements are false.

3. Statement 1 is true, and statement 2 is false.

4. Statement 1 is false, and statement 2 is true.

Answer

Since AB = AC, ΔABC is an isosceles triangle.

The angles opposite to equal sides of a triangle are equal.

⇒ ∠B = ∠C

∠ADB is an exterior angle to ΔADC.

Therefore, ∠ADB > ∠C

⇒ ∠ADB > ∠B

⇒ AB > AD (Side opposite to the largest angle in a triangle is the longest side)

∴ Both the statements are true.

Hence, option 1 is the correct option.

Assertion (A): In the given figure, AB = BC and AD = CE. Then BD will be equal to BE.

Reason (R): ΔBAD ≅ ΔBCE by SAS.

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true and R is the correct reason for A.

4. Both A and R are true and R is the incorrect reason for A.

Answer

It is given that AB = BC.

⇒ ∠BAC = ∠BCA [Angles opposite to equal sides of the triangle are also equal]

⇒ ∠BAD = ∠BCE ..................(1)

In ΔBAD and ΔBCE,

⇒ AB = BC [Given]

⇒ AD = CE [Given]

⇒ ∠BAD = ∠BCE

∴ ΔBAD ≅ ΔBCE (By SAS congruency criterion)

So, reason (R) is true.

By C.P.C.T.,

BE = BD

So, assertion (A) is true.

∴ Both A and R are true and R is the correct reason for A.

Hence, option 3 is the correct option.

Assertion (A): In the given figure, S is any point on side QR.

∴ PQ + QR + RP > 2PS

Reason (R): In ΔPQS, PQ + QS > PS and in ΔPRS, PR + SR > PS

1. A is true, but R is false.

2. A is false, but R is true.

3. Both A and R are true and R is the correct reason for A.

4. Both A and R are true and R is the incorrect reason for A.

Answer

In △ PQS,

⇒ PQ + QS > PS ......(1) [Sum of any two sides of a triangle is greater than the third side]

In △ PRS,

⇒ RP + RS > PS ......(2) [Sum of any two sides of a triangle is greater than the third side]

So, reason (R) is true.

Adding equations (1) and (2), we get :

⇒ PQ + QS + RP + RS > PS + PS

⇒ PQ + (QS + RS) + RP > 2PS

⇒ PQ + QR + RP > 2PS.

So, assertion (A) is true.

∴ Both A and R are true, and R is the correct reason for A.

Hence, option 3 is the correct option.

In the figure given alongside, AD = AB = AC, BD is parallel to CA and angle ACB = 65°. Find angle DAC.

Answer

From figure,

In △ ABC,

⇒ AB = AC (Given)

⇒ ∠ABC = ∠ACB = 65° (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠ABC + ∠ACB + ∠BAC = 180°

⇒ 65° + 65° + ∠BAC = 180°

⇒ 130° + ∠BAC = 180°

⇒ ∠BAC = 180° - 130° = 50°.

Given,

BD || CA

∴ ∠DBA = ∠BAC = 50° (Alternate angles are equal)

In △ DAB,

⇒ AD = AB (Given)

⇒ ∠ADB = ∠DBA = 50° (Angles opposite to equal sides are equal)

By angle sum property of triangle,

⇒ ∠ADB + ∠DBA + ∠DAB = 180°

⇒ 50° + 50° + ∠DAB = 180°

⇒ 100° + ∠DAB = 180°

⇒ ∠DAB = 180° - 100° = 80°.

From figure,

⇒ ∠DAC = ∠DAB + ∠BAC = 80° + 50° = 130°.

Hence, ∠DAC = 130°.

Prove that a triangle ABC is isosceles, if :

(i) altitude AD bisects angle BAC or,

(ii) bisector of angle BAC is perpendicular to base BC.

Answer

Let AD be the altitude on side BC.

We know that,

Altitude from a point is always perpendicular to other side.

Let altitude AD bisect angle BAC.

Hence, segment AD satisfies both the conditions.

In △ ADB and △ ADC,

⇒ AD = AD (Common side)

⇒ ∠BAD = ∠CAD (Since, AD bisects angle BAC)

⇒ ∠ADB = ∠ADC (Since, AD is altitude to side BC)

∴ Δ ADB ≅ Δ ADC (By A.S.A. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

⇒ AB = AC.

Hence, proved that ABC is an isosceles triangle.

In △ABC, angle ∠ACB = 90°. D is a point on side AB so that DA = DC.

(i) Prove that △BDC is an isosceles triangle.

(ii) If ∠BDC = 60°, show that ∠A = 30°.

Answer

(i) In △ADC,

Since AD = DC, △ADC is an isosceles triangle.

In isosceles triangle, angles opposite to equal sides are equal.

∴ ∠DAC = ∠DCA.

Let ∠DCA be 'x'.

In △ABC,

∠ACB = 90°

⇒ ∠ACB = ∠DCA + ∠DCB

⇒ ∠DCB = ∠ACB - ∠DCA

⇒ ∠DCB = 90° - x ........(1)

By angle sum property of triangle,

In triangle ABC,

⇒ ∠ACB + ∠A + ∠B = 180°

⇒ ∠B = 180° - (∠ACB + ∠A)

⇒ ∠B = 180° - (90° + x)

⇒ ∠B = 90° - x ........(2)

From equation (1) & (2), we get :

∠DCB = ∠B

Since, ∠DCB = ∠B, the sides opposite to them must be equal (CD = BD)

∴ △BDC is an isosceles triangle.

Hence, proved that △BDC is an isosceles triangle.

(ii) Given, ∠BDC = 60°

From figure,

∠ADC + ∠BDC = 180°

∠ADC + 60° = 180°

∠ADC = 120°

Since, triangle ADC is an isosceles triangle, with AD = CD.

Thus, ∠DAC = ∠DCA = a (let)

By angle sum property of triangle,

∠DAC + ∠DCA + ∠ADC = 180°

a + a + 120° = 180°

2a = 180° - 120°

2a = 60°

a = $\dfrac{60\degree}{2}$ = 30°.

Hence proved that ∠A = 30°.

In the following figure; IA and IB are bisectors of angles CAB and CBA respectively. CP is parallel to IA and CQ is parallel to IB.

Prove that :

PQ = The perimeter of Δ ABC.

Answer

From figure,

IA || CP and CA is a transversal.

⇒ ∠CAI = ∠PCA (Alternate angles are equal) .........(1)

Also, IA || CP and AP is a transversal

⇒ ∠IAB = ∠APC (Corresponding angles are equal) ......(2)

Since, IA is the bisector of ∠CAB.

∴ ∠CAI = ∠IAB .........(3)

Substituting values from equation (1) and (2) in equation (3), we get :

⇒ ∠PCA = ∠APC

⇒ AC = AP (Sides opposite to equal angles are equal)

From figure,

IB || CQ and CB is a transversal.

⇒ ∠CBI = ∠QCB (Alternate angles are equal) .........(4)

Also, IB || CQ and BQ is a transversal

⇒ ∠IBA = ∠BQC (Corresponding angles are equal) ......(5)

Since, IB is the bisector of ∠CBA.

∴ ∠CBI = ∠IBA .........(6)

Substituting values from equation (4) and (5) in equation (6), we get :

⇒ ∠QCB = ∠BQC

⇒ BQ = BC (Sides opposite to equal angles are equal)

From figure,

PQ = AP + AB + BQ = AC + AB + BC

= Perimeter of △ ABC.

Hence, proved that PQ = Perimeter of △ ABC.

The given figure shows an equilateral triangle ABC with each side 15 cm. Also DE // BC, DF // AC and EG //AB. If DE + DF + EG = 20 cm, find FG.

Answer

Given,

ABC is an equilateral triangle.

∴ AB = BC = AC = 15 cm and ∠A = ∠B = ∠C = 60°.

In △ ADE,

⇒ ∠ADE = ∠ABC = 60° (Corresponding angles are equal)

⇒ ∠AED = ∠ACB = 60° (Corresponding angles are equal)

By angle sum property of triangle,

⇒ ∠AED + ∠ADE + ∠DAE = 180°

⇒ 60° + 60° + ∠DAE = 180°

⇒ ∠DAE + 120° = 180°

⇒ ∠DAE = 180° - 120° = 60°.

∴ △ ADE is an equilateral triangle with each side equal to x cm.

∴ AD = DE = EA = x cm.

In △ BDF,

⇒ ∠BFD = ∠BCA = 60° (Corresponding angles are equal)

⇒ ∠DBF = ∠B = 60°

By angle sum property of triangle,

⇒ ∠BFD + ∠DBF + ∠BDF = 180°

⇒ 60° + 60° + ∠BDF = 180°

⇒ ∠BDF + 120° = 180°

⇒ ∠BDF = 180° - 120° = 60°.

∴ △ BDF is an equilateral triangle with each side equal to y cm.

∴ DB = BF = FD = y cm.

In △ EGC,

⇒ ∠EGC = ∠ABC = 60° (Corresponding angles are equal)

⇒ ∠ECG = ∠C = 60°

By angle sum property of triangle,

⇒ ∠EGC + ∠ECG + ∠GEC = 180°

⇒ 60° + 60° + ∠GEC = 180°

⇒ ∠GEC + 120° = 180°

⇒ ∠GEC = 180° - 120° = 60°.

∴ △ EGC is an equilateral triangle with each side equal to z cm.

∴ EG = GC = CE = z cm.

Given,

⇒ AB = 15

⇒ AD + BD = 15

⇒ x + y = 15 ..........(1)

⇒ AC = 15

⇒ AE + EC = 15

⇒ x + z = 15 ..........(2)

Given,

⇒ DE + DF + EG = 20

⇒ x + y + z = 20

⇒ 15 + z = 20 [From equation (1)]

⇒ z = 20 - 15 = 5 cm.

Substituting value of z in equation (2), we get :

⇒ x + 5 = 15

⇒ x = 15 - 5 = 10 cm.

Substituting value of x in equation (1), we get :

⇒ 10 + y = 15

⇒ y = 15 - 10 = 5 cm.

From figure,

⇒ BC = 15 cm

⇒ BF + FG + GC = 15

⇒ y + FG + z = 15

⇒ 5 + FG + 5 = 15

⇒ FG + 10 = 15

⇒ FG = 15 - 10 = 5 cm.

Hence, FG = 5 cm.

In triangle ABC, bisector of angle BAC meets opposite side BC at point D. If BD = CD, prove that △ ABC is isosceles.

Answer

Produce AD upto E such that AD = DE.

In △ ABD and △ EDC,

⇒ AD = DE (Given)

⇒ BD = CD (Given)

⇒ ∠ADB = ∠EDC (Vertically opposite angles are equal)

∴ △ ABD ≅ △ EDC (By S.A.S. axiom)

We know that,

Corresponding parts of congruent triangle are equal.

⇒ AB = CE .........(1)

⇒ ∠BAD = ∠CED

⇒ ∠BAD = ∠CAD (As AD is the bisector BAC)

∴ ∠CAD = ∠CED

∴ CE = AC (Sides opposite to equal angles are equal) ..........(2)

From equations (1) and (2), we get :

⇒ AB = AC.

Hence, proved that ABC is an isosceles triangle.

In Δ ABC, D is a point on BC such that AB = AD = BD = DC. Show that :

∠ADC : ∠C = 4 : 1.

Answer

Given,

AB = AD = BD

∴ Δ ABD is an equilateral triangle.

∴ ∠ABD = ∠ADB = ∠BAD = 60°.

Since, BDC is a straight line.

∴ ∠ADB + ∠ADC = 180°

⇒ 60° + ∠ADC = 180°

⇒ ∠ADC = 180° - 60° = 120°.

In Δ ADC,

⇒ AD = DC (Given)

∴ ∠DAC = ∠DCA = x (let) [Angles opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠DAC + ∠DCA + ∠ADC = 180°

⇒ x + x + 120° = 180°

⇒ 2x = 180° - 120°

⇒ 2x = 60°

⇒ x = $\dfrac{60°}{2}$ = 30°

⇒ ∠DCA = ∠C = 30°.

⇒ ∠ADC : ∠C = 120° : 30° = 4 : 1.

Hence, proved that ∠ADC : ∠C = 4 : 1.

Using the information, given in each of the following figures, find the values of a and b.

Answer

In △ CAE,

⇒ AC = CE (Given)

∴ ∠AEC = ∠EAC = x (let) [Angles opposite to equal sides are equal]

By angle sum property of triangle,

⇒ ∠AEC + ∠EAC + ∠ACE = 180

⇒ x + x + 68° = 180°

⇒ 2x = 180° - 68°

⇒ 2x = 112°

⇒ x = $\dfrac{112°}{2}$ = 56°

⇒ ∠AEC = ∠EAC = 56°.

Since, B, E and C are in a straight line.

∴ ∠BEA + ∠AEC = 180°

⇒ a + 56° = 180°

⇒ a = 180° - 56° = 124°.

In △ ABE,

By angle sum property of triangle,

⇒ ∠ABE + ∠BEA + ∠EAB = 180°

⇒ ∠ABE + 124° + 14° = 180°

⇒ ∠ABE + 138° = 180°

⇒ ∠ABE = 180° - 138° = 42°.

Since, AB || CD

⇒ b = ∠ABE = 42°. (Alternate angles are equal)

Hence, a = 124° and b = 42°.

Using the information, given in each of the following figures, find the values of a and b.

Answer

In △ ADE,

⇒ AD = AE (Given)

∴ ∠ADE = ∠AED (Angles opposite to equal sides are equal)

⇒ 180° - ∠ADE = 180° - ∠AED

⇒ ∠ADC = ∠AEB

In △ ABE and △ CAD,

⇒ ∠EAB = ∠CAB (Given)

⇒ ∠ADC = ∠AEB (Proved above)

⇒ AE = AD (Given)

∴ △ ABE ≅ △ CAD (By A.S.A. axiom)

We know that,

Corresponding sides of congruent triangle are equal.

∴ AC = AB and CD = EB.

Considering CD = EB,

⇒ a = 3b .........(1)

Considering AC = AB,

⇒ 2a + 2 = 7b - 1

Substituting value of a from equation (1) in above equation, we get :

⇒ 2(3b) + 2 = 7b - 1

⇒ 6b + 2 = 7b - 1

⇒ 7b - 6b = 2 + 1

⇒ b = 3.

Substituting value of b in equation (1), we get :

⇒ a = 3(3) = 9.

Hence, a = 9 and b = 3.

In the following figure; AB is the largest side and BC is the smallest side of triangle ABC.

Write the angles x°, y° and z° in ascending order of their values.

Answer

Given,

AB is the largest side and BC is the smallest side of triangle ABC.

∴ AB > AC > BC

⇒ ∠C > ∠B > ∠A (If two sides of a triangle are unequal, the greater side has the greater angle opposite to it.)

⇒ 180° - ∠C < 180° - ∠B < 180° - ∠A

⇒ z° < y° < x°.

Hence, z° < y° < x°.

In quadrilateral ABCD, side AB is the longest and side DC is the shortest. Prove that :

(i) ∠C > ∠A

(ii) ∠D > ∠B

Answer

(i) Given,

In quadrilateral ABCD,

AB is the longest sides and DC is the shortest side.

Join BD and AC.

In △ ABC,

⇒ AB > BC

∴ ∠1 > ∠2 .......(1) [If two sides of a triangle are unequal, the greater side has the greater angle opposite to it.]

In △ ADC,

⇒ AD > DC

∴ ∠7 > ∠4 .......(2) [If two sides of a triangle are unequal, the greater side has the greater angle opposite to it.]

Adding equations (1) and (2), we get :

⇒ ∠1 + ∠7 > ∠2 + ∠4

⇒ ∠C > ∠A.

Hence, proved that ∠C > ∠A.

(ii) In △ ABD,

⇒ AB > AD

∴ ∠5 > ∠6 .......(1) [If two sides of a triangle are unequal, the greater side has the greater angle opposite to it.]

In △ BDC,

⇒ BC > CD

∴ ∠3 > ∠8 .......(2) [If two sides of a triangle are unequal, the greater side has the greater angle opposite to it.]

Adding equations (1) and (2), we get :

⇒ ∠5 + ∠3 > ∠6 + ∠8

⇒ ∠D > ∠B.

Hence, proved that ∠D > ∠B.

In triangle ABC, side AC is greater than side AB. If the internal bisector of angle A meets the opposite side at point D, prove that : ∠ADC is greater than ∠ADB.

Answer

In △ ADC,

⇒ ∠ADB = ∠1 + ∠C [In a triangle an exterior angle is equal to the sum of two opposite interior angles.] .........(1)

In △ ADB,

⇒ ∠ADC = ∠2 + ∠B [In a triangle an exterior angle is equal to the sum of two opposite interior angles.] .........(2)

In △ ABC,

⇒ AC > AB (Given)

⇒ ∠B > ∠C [If two sides of a triangle are unequal, the greater side has the greater angle opposite to it.]

Since, AD is the bisector of angle A.

∴ ∠2 + ∠B > ∠1 + ∠C .........(3)

From equations (1), (2) and (3), we get :

⇒ ∠ADC > ∠ADB.

Hence, proved that ∠ADC is greater than ∠ADB.

In isosceles triangle ABC, sides AB and AC are equal. If point D lies in base BC and point E lies on BC produced (BC being produced through vertex C), prove that :

(i) AC > AD

(ii) AE > AC

(iii) AE > AD

Answer

(i) In ∆ ABC,

⇒ AB = AC (Given)

⇒ ∠ACB = ∠ABC [Angles opposite to equal sides are equal] .......(1)

⇒ ∠ACD = ∠ABD [From figure, ∠ACB = ∠ACD and ∠ABC = ∠ABD] .........(2)

We know that exterior angle of a triangle is always greater than each of the interior opposite angle.

In ∆ ADC,

⇒ ∠ADB > ∠ACD

⇒ ∠ADB > ∠ABD [Using equation (2)]

⇒ AB > AD [If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.]

⇒ AC > AD [As, AB = AC] .....(3)

Hence, proved that AC > AD.

(ii) We know that exterior angle of a triangle is always greater than each of the interior opposite angle.

In ∆ ACE,

⇒ ∠ACD > ∠AEC ........(4)

From equation (2),

⇒ ∠ACD = ∠ABD

From figure,

⇒ ∠ABD = ∠ABE

⇒ ∠ACD = ∠ABE

⇒ ∠AEC = ∠AEB

Substituting value of ∠ACD and ∠AEC in equation (4), we get :

⇒ ∠ABE > ∠AEB

In △ AEB,

⇒ AE > AB [If two angles of a triangle are unequal, the greater angle has the greater side opposite to it.]

As, AB = AC,

∴ AE > AC

Hence, proved that AE > AC.

(iii) Since, AE > AC and AC > AD,

∴ AE > AD.

Hence, proved that AE > AD.

Given : ED = EC

Prove : AB + AD > BC.

Answer

We know that,

The sum of any two sides of the triangle is always greater than the third side of the triangle.

In △ CEB,

⇒ EC + EB > BC

⇒ ED + EB > BC (As, EC = ED)

⇒ BD > BC

In △ ADB,

⇒ AD + AB > BD

Since, BD > BC and AD + AB > BD

∴ AD + AB > BC.

Hence, proved that AB + AD > BC.