Question 1: If two isosceles triangles have a  common base, prove that the line joining their vertices bisects them at right angles.

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.

$\\$

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 $\angle BDC$

(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 $BC$

$\\$

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.

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

$\\$

Question 4: In the adjoining figure, it is given that $AB = CD$ and $AC = BD$. Prove that $\triangle ADC \cong \triangle CBA$

Consider $\triangle ABC$ and $\triangle ADC$

$AB = DC$ (given)

$AC = BD$ (given)

$AD$ is common

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

$\\$

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 $DF = EF$

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)

$\therefore DE = FE$

$\\$

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.

Consider $\triangle BDF$ and $\triangle CEF$

$BF = FC$ (given)

$FD = FE$ (given)

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

$\therefore \triangle BDF \cong CEF$

$\Rightarrow \angle DBF = \angle ECF$

$\Rightarrow AB = AC$

$\\$

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 $\angle BAF = \angle ABE$

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 $\angle BAF = \angle ABE$

$\\$

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 \$.

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 $\angle ACD = \angle DCB$

$\therefore DB > DC$

$\\$

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 $AB > AC$

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 $x > y$

$\therefore \angle ABC < \angle ACB$

$\therefore AB > AC$

$\\$

Question 10: If $D$ is any point on the base $BC$ produced, of an isosceles triangle $ABC$, prove that $AD > AB$

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

$\angle ACB > \angle ADB$

$\Rightarrow \angle ACD > \angle ADC$

$\Rightarrow AD > AC$

$\Rightarrow AD > AB$

$\\$

Question 11: In the adjoining figure, if $AD$ is the bisector of $\angle BAC$, show that (i) $AB > BD$ (ii) $AC > CD$

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 $AC < CD$.

$\\$

Question 12: In the adjoining figure, $AC > AB$ and $AD$ is the bisector of  $\angle BAC$. Show that $\angle ADC > \angle ADB$

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 $\angle ADC > \angle ADB$

$\\$

Question 13: Show that the sum of three altitudes of a triangle is less than the sum of the three sides of the triangle.

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

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

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

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

$\\$

Question 14: Prove that the perimeter of a triangle is greater than the sum of its three medians.

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

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

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

$\\$

Question 15: In the adjoining figure, $ABC$ is a triangle and $D$ is any point in its interior, show that $DB + DC < AB + AC$

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,

$AB + AE + DE + EC > DB + DE + DC$

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

$\Rightarrow AB + AC > DB + DC$

$\Rightarrow DB + DC < AB + AC$

$\\$

Question 16: In the adjoining figure, $AP \perp l$ and $PR > PQ$. Show that $AR > AQ$

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 $AC > AB$

$\\$

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 $\angle ADC > \angle ABC$

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.

$\\$

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) $AB + BC + CD + DA < 2(AC + BD)$

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 $OB + OD = BD$

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

$\\$

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

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

$OA + OB > AB$

$OB + OC > BC$

$OC + OA > AC$

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

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

$\\$

Question 20: In the adjoining  figure,  prove that: (i) $CD + DA + AB + BC > 2 AC$ (ii) $CD + DA + AB > BC$

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 $AB + AC > BC$)

$\Rightarrow CD + DA + AB > BC$

$\\$