Question 1: If two isosceles triangles have a  common base, prove that the line joining their vertices bisects them at right angles.2018-09-02_10-06-13

Answer:

Given: AB = AC and DB = DC 

To prove: AD bisects BC at right angles

Proof: In \triangle ABD and \triangle ACD we have

AB = AC (Given)

BD = CD (Given)

and AD is common

Therefore \triangle ABD \cong \triangle ACD (By SSS theorem))

\Rightarrow \angle BAE = \angle EAC

Thus in \triangle ABE and \triangle ACE , we have

AB = AC (Given) and \angle BAE = \angle EAC (By SAS theorem)

\Rightarrow BE = CE and \angle AEB = \angle AEC 

But we know, \angle AEB + \angle AEC = 190^o

\Rightarrow \angle AEB = \angle AEC = 90^o

Hence, AD bisects BC at right angles.

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Question 2: \triangle ABC and \triangle DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC . If AD is extended to intersect BC at E , show that (i) \triangle ABD \cong \triangle ACD (ii) \triangle ABE \cong \triangle ACE  (iii) AE bisect \angle BAC as well as \angle BDC
2018-09-02_10-06-38

Answer:

(i)   In \triangle ABD and \triangle ACD , we have

AB = AC (Given)

BD = CD (Given)

and AD is common

Therefore \triangle ABD \cong \triangle ACD (By SSS theorem)

(ii) In \triangle ABE and \triangle ACE 

AB = AC  (Given)

\angle BAE = \angle CAE (corresponding angles of congruent triangles are equal)

and AE is common.

Therefore, \triangle ABE \cong \angle ACE (by SAS theorem)

(iii) In (i) we proved \triangle ABD \cong \triangle ACD

\Rightarrow \angle BAD = \angle CAD

\Rightarrow \angle BAE = \angle CAE

\Rightarrow AE is the bisector of \angle BAC

In \triangle BDE and \triangle CD we have

BD = CD (Given)

BE = CE (corresponding sides of congruent triangles)

and DE is common

Therefore \triangle BDE \cong \triangle CDE (By SSS theorem)

\Rightarrow \angle BDE = \angle CDE

\Rightarrow DE is the bisector of \angle BDC

Hence AE is the bisector of \angle BAC as well as \angle BDC

(iv) In (iii) we proved that, \triangle BDE \cong \triangle CDE

\Rightarrow BE = CE and \angle BED = \angle CED

\Rightarrow BE = CE and \angle BED = \angle CED = 90^o

\Rightarrow DE is the perpendicular bisector of BC

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Question 3:  A point E is taken inside an equilateral four sides figure ABCD such that its distances from the angular points D and B are equal. Show that AE and EC are in one and in the same straight line. 2018-09-02_10-06-49

Answer:

Given: BE = ED

To Prove: AE and EC are in one and the same straight line

Proof: In \triangle AED and \triangle AEB , we have

AD = AB (given)

AE is common

ED = EB (given)

\therefore \triangle AED \cong \triangle AEB (by SSS theorem)

\Rightarrow \angle AEB = \angle AED

Similarly, \triangle DEC \cong \triangle BEC

\Rightarrow \angle DEC = \angle CEB

But \angle AEB + \angle AED + \angle DEC + \angle CEB = 4 right angles

\Rightarrow 2 \angle AED + 2 \angle DEC = 360^o

\Rightarrow \angle AED + \angle DEC = 180^o

\Rightarrow \angle AED and \angle DEC form a linear pair

\Rightarrow AE and EC are in same straight line

\Rightarrow AC is a straight line

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Question 4: In the adjoining figure, it is given that AB = CD and AC = BD . Prove that \triangle ADC \cong \triangle CBA 2018-09-02_10-07-02

Answer:

Consider \triangle ABC and \triangle ADC

AB = DC (given)

AC = BD (given)

AD is common

\therefore \triangle ABC \cong \triangle ADC (By SSS theorem)

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Question 5:  In \triangle ABC , if AB = AC and D, E and F are the midpoints of the sides AB, BC and AC respectively. Prove that DF = EF 2018-09-02_10-07-19

Answer:

In \triangle ABC

AB = AC (given)

AD = DB, BF = FC and AE = EC (given)

To prove: DE = FE

Using the Mid Point theorem

FE \parallel AB and FE = \frac{1}{2} AB

Similarly DF = AE

In \triangle EFD and \triangle DAE

FE = AD

DF = AE

DE is common

\therefore \triangle EFD \cong \triangle DAE (by SSS theorem)

\therefore DE = FE

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Question 6: In the adjoining figure, it is given that DF = FE, BF = FC and FD \perp AB and FE \perp AC . Prove that AB = AC . Hence prove that the triangles is isosceles. 2018-09-02_10-07-33

Answer:

Consider \triangle BDF and \triangle CEF

BF = FC (given)

FD = FE (given)

and \angle FDB = \angle FEC = 90^o

\therefore \triangle BDF \cong CEF

\Rightarrow \angle DBF = \angle ECF

\Rightarrow AB = AC

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Question 7: ABCD is a square, E and F are points on the side AD and BC respectively such that AF = BE . Prove that BF = AE and \angle BAF = \angle ABE 2018-09-02_10-07-45

Answer:

Consider \triangle ABE and \triangle ABF

AB is common

AE = AF (given)

\angle EAB = \angle ABF = 90^o

\therefore \triangle ABE \cong \triangle ABF

\therefore AE = BF

and \angle BAF = \angle ABE

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Question 8: In the adjoining figure, AB > AC . BD and CD are the bisectors of \angle ABC and \angle ACB respectively. Prove that DB latex > DC $. 2018-09-02_10-08-15

Answer:

In \triangle ABC

AB > AC (given)

\Rightarrow \angle ACB > \angle ABC

(angle opposite to the larger side of a triangle is greater)

\Rightarrow \frac{1}{2} \angle ACB = \frac{1}{2} \angle ABC

\Rightarrow \angle DCB = \angle DBC

and \angle ACD = \angle DCB

\therefore DB > DC

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Question 9: In the adjoining figure, side AB and AC of  \triangle ABC are extended to D and E respectively. If x > y , show that AB > AC 2018-09-02_10-08-28

Answer:

We have

\angle ABC + x = 180^o (angles of a linear pair)

\angle ACB + y = 180^o (angles of a linear pair)

\therefore \angle ABC + x = \angle ACB + y

Since x > y

\therefore \angle ABC < \angle ACB

\therefore AB > AC

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Question 10: If D is any point on the base BC produced, of an isosceles triangle ABC , prove that AD > AB 2018-09-02_10-08-45

Answer:

In \triangle ABC

AB = BC (given)

\Rightarrow \angle ABC = \angle ACB (angles opposite equal sides of a triangle are equal)

In \triangle ABD , we have

Exterior \angle ABC > \angle ADB … … … … … i)

(since exterior angle of a triangle is greater than each of interior opposite angles)

\Rightarrow \angle ABC > \angle ADB … … … … …. ii)

From i) and ii), we get

\angle ACB > \angle ADB

\Rightarrow \angle ACD > \angle ADC

\Rightarrow AD > AC

\Rightarrow AD > AB

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Question 11: In the adjoining figure, if AD is the bisector of \angle BAC , show that (i) AB > BD (ii) AC > CD  2018-09-02_10-09-08

Answer:

In \triangle ABC ,

AD is the bisector of \angle BAC

\angle BAD = \angle DAC (since exterior angle of a triangle is greater than each of interior opposite angles)

Therefore in \triangle ADC

Ext. \angle ADC > \angle DAC

\Rightarrow \angle BDA > \angle DAC

\Rightarrow \angle BDA > \angle BAD

Thus in \triangle ABD we have

\angle BDA > \angle BAD

\Rightarrow AB > BD (sides opposite to the greater angle is larger)

Hence AB > BD .

Similarly, we can prove AC < CD .

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Question 12: In the adjoining figure, AC > AB and AD is the bisector of  \angle BAC . Show that \angle ADC > \angle ADB 2018-09-02_10-09-24

Answer:

In \triangle ABC , we have

AC > AB

\Rightarrow \angle ABC > \angle ACB

\Rightarrow \angle ABC + \angle BAD > \angle ACB + \angle BAD

\Rightarrow \angle ABC + \angle BAD > \angle ACB + \angle DAC

Since \angle BAD = \angle DAC

Now in \triangle ABD and \triangle ADC we have

\angle ABC + \angle BAD + \angle ADB = 180^o

\angle ACB + \angle DAC + \angle ADC = 180^o

\angle ABC + \angle BAD = 180^o - \angle ADB

and \angle ACB + \angle DAC = 180^o - \angle ADC

\therefore 180^o - \angle ADB > 180^o - \angle ADC

180^o - \angle ADB - 180^o - \angle ADC > 0

\angle ADC - \angle ADB  > 0

or \angle ADC > \angle ADB

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Question 13: Show that the sum of three altitudes of a triangle is less than the sum of the three sides of the triangle. 2018-09-02_10-09-36

Answer:

Given \angle ADC = \angle BEA = \angle CFA = 90^o (altitudes)

AD \perp BC

\Rightarrow AD > AD and AC > AD

\Rightarrow AB + AC > AD + AD

\Rightarrow AB + AC > 2 AD … … … … … i)

Similarly, BE \perp AC

\Rightarrow BC > BE and BA > BE

\Rightarrow BC + BA > 2 BE … … … … … ii)

and CF \perp AB

\Rightarrow AC > CF and BC > CF

\Rightarrow AC + BC > 2CF   … … … … … iii)

Adding i) , ii) and iii)  we get

(AB + AC) + (AB + BC) + (AC + BC) > 2 AD + 2 BE + 2CF

\Rightarrow 2(AB + BC + AC) > 2 (AD + BE + CF)

\Rightarrow AB + BC + AC > 2 AD + BE + CF

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Question 14: Prove that the perimeter of a triangle is greater than the sum of its three medians. 2018-09-02_10-09-57

Answer:

In \triangle ABC, AD, BE and CF are medians.

To prove: AB + BC + CF > AD + BE + CF

Proof: We know that the sum of any two sides of a triangle is greater then the twice of the median bisecting the third side.

AD is the median bisecting BC

\Rightarrow AB + AC > 2AD

Similarly, AB + BC > 2BE

and BC + AC > 2CF

Adding i), ii) and iii) we get

(AB + AC) + (AB + BC) + (BC + AC) > 2AD + 2BE +2CF

\Rightarrow AB + BC + AC > AD + BE + CF

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Question 15: In the adjoining figure, ABC is a triangle and D is any point in its interior, show that DB + DC < AB + AC  2018-09-02_10-10-10

Answer:

Given D is any point inside \triangle ABC

To Prove: DB + DC < AB + AC

Proof: In \triangle ABE we have

AB + AE > BE   (Since the sum of two sides of a triangle is greater than the third side)

\Rightarrow AB + AE > BD + DE … … … … … i)

In \triangle CDE , we have

DE + EC > DC … … … … … ii)

Adding i) and ii) we get,

AB + AE + DE + EC > DB + DE + DC 

\Rightarrow AB + (AE + EC) > DB + DC

\Rightarrow AB + AC > DB + DC

\Rightarrow DB + DC < AB + AC

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Question 16: In the adjoining figure, AP \perp l and PR > PQ . Show that AR > AQ 2018-09-02_10-10-21

Answer:

Given AC \perp AB

In \triangle ADB and \triangle ADE , we have

AD is common

\angle ADB = \angle ADE = 90^o

and DB = DE (by construction)

\therefore \triangle ADB \cong \triangle ADE (by SAS theorem)

\Rightarrow AB = AE (since corresponding part of congruent triangles)

Thus in \triangle ABE , we have AB = AE

\angle ABD = \angle AED   … … … … … i) (angles opposite equal sides are equal)

In \triangle ACE , we have

\angle AED > \angle ACD … … … … … ii)

from i) and ii) we get

\angle ABD < \angle ACD

\Rightarrow AC > AB (since side opposite to greater angle is larger)

Hence AC > AB

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Question 17:  In the adjoining figure, ABCD is a quadrilateral. AB is the longest side and CD is the shortest side. Prove that \angle DCB > \angle DAB and \angle ADC > \angle ABC 2018-09-02_10-10-30

Answer:

To prove: i) \angle C > \angle A ii) \angle D > \angle B

Construction: Join AC and BD

Since AB is the longest side of Quadrilateral ABCD , therefore in \triangle ABC , we have

AB > BC

\Rightarrow \angle ACB > \angle CAB … … … … … i)

Since CD is the shortest side of the quadrilateral ABCD , therefore we have in \triangle ADC

AD > CD

\Rightarrow \angle ACD > \angle CAD … … … … … ii)

Adding i) and ii) we get

\angle ACB + \angle ACD > \angle CAB + \angle CAD

\Rightarrow \angle C > \angle A

ii) In \triangle ABD , we have

AB > AD

\Rightarrow \angle ADB > \angle DBA   … … … … … iii)

In \triangle DCB , we have

CB > CD

\Rightarrow \angle BDC > DBC   … … … … … iv)

Adding iii) and iv) we get

\angle ADB + \angle BDC > \angle DBA + \angle DBC

\Rightarrow \angle D > \angle B

Hence Proved.

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Question 18: In the adjoining figure, ABCD is a quadrilateral in which diagonals AC and BD intersect at O . Show that (i) AB + BC + CD + DA > AC + BD (ii) AB + BC + CD + DA < 2(AC + BD)  2018-09-02_10-10-41

Answer:

i) Since the sum of any two sides of a triangle is greater than the third side, therefore

In \triangle ABC we have

AB + BC > AC … … … … … i)

In \triangle CDA we have

CD + DA > AC … … … … … ii)

In \triangle ABD , we have

AB + DA > BD … … … … … iii)

In \triangle BCD , we have

BC + CD > BD … … … … … iv)

Adding i), ii), iii) and iv) we get

2(AB + BC + CD + DA) > 2(AC + BD)

\Rightarrow AB + BC + CD + DA > AC + BD

In \triangle OAB , we have

OA + OB > AB … … … … … v)

In \triangle OBC , we have

OB + OC > BC … … … … … vi)

In \triangle OCD , we have

OC + OD > CD … … … … … vii)

In \triangle ODA , we have

OD + OA > DA … … … … … viii)

Adding v), vi), vii) and viii) we get

2(OA+OB+OC+OD) > AB + BC + CD + DA

\Rightarrow 2[(OA + OC) + (OB+OD)] >  AB + BC + CD + DA

2(AC+BD)>  AB + BC + CD + DA

Since OA + OC = AC and OB + OD = BD

\Rightarrow  AB + BC + CD + DA < 2(AC - BD)

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Question 19: O is any point in the interior of \triangle ABC . Prove that (i) AB + AC > OB + OC    (ii) AB + BC + CA > OA + OB +OC (iii) OA + OB + OC > \frac{1}{2} (AB + BC + CA)  2018-09-02_10-10-56

Answer:

In \triangle ABD we have

AB + AD > BD

\Rightarrow AB + AD > OB + OD … … … … … i)

In \triangle ODC , we have

OD + DC > OC … … … … … ii)

Adding i) and ii), we get

AB + AD + OD + DC > OB + OD + OC

\Rightarrow AB + AC > OB + OC … … … … … iii)

ii) Similarly,

BC + BA > OA + OC … … … … … iv)

and CA + CB > OA + OB … … … … … v)

Adding iii) , iv) and v) we get

2(AB + BC + CA) > 2 (OA + OB + OC)

\Rightarrow AB + BC + CA >OA + OB + OC

In \triangle OAB, \triangle OBC and \triangle OCA , we have

OA + OB > AB

OB + OC > BC

OC + OA > AC

\Rightarrow 2(OA + OB + OC) > AB + BC + C

\Rightarrow OA + OB + OC > \frac{1}{2} (AB + BC + CA)

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Question 20: In the adjoining  figure,  prove that: (i) CD + DA + AB + BC > 2 AC (ii) CD + DA + AB > BC 2018-09-02_10-11-06

Answer:

i) In \triangle ABC we have

AB + BC > AC … … … … … i)

In \triangle ACD we have

AD + CD > AC … … … … … ii)

Adding i) and ii) we get

AB + BC + AD + CD >2 AC

ii) In \triangle ACD we have

CD + DA > CA

\Rightarrow CD + DA + AB > CA + AB (since AB + AC > BC )

\Rightarrow CD + DA + AB > BC

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