##### Question 1:

In quadrilateral ACBD, AC = AD and AB bisects ∠A (see fig.). Show that $\u2206$ABC $\cong $ $\u2206$ABD.

What can you say about BC and BD?

##### Answer:

Given that, in quadrilateral ACBD;

AC = AD and AB bisects ∠A.

We have to prove that $\u2206$ABC $\cong $ $\u2206$ABD.

Proof: In $\u2206$ABC and $\u2206$ABD,

AC = AD (given)

∠BAC = ∠BAD [$\because $ AB bisects ∠A (given)]

AB = AB (common)

$\therefore $ $\u2206$ABC $\cong $ $\u2206$ABD [SAS criteria of congruency]

Thus BC = BD

(Congruent parts of congruent triangles)

##### Question 2:

ABCD is a quadrilateral in which AD = BC

and ∠DAB = ∠CBA (see fig.). Prove that

(i ) $\u2206$ABD $\cong $ $\u2206$BAC

(ii) BD = AC

(iii) ∠ABD = ∠BAC.

##### Answer:

In $\u2206$ABD and $\u2206$ABC,

AD = BC (given)

∠DAB = ∠CBA (given)

AB = AB (common)

(i ) $\u2206$ABD $\cong $ $\u2206$BAC

(SAS criteria of congruency)

(ii) BD = AC

(Congruent parts of congruent triangles)

(iii) ∠ABD = ∠BAC

(Congruent parts of congruent triangles)

##### Question 3:

AD and BC are equal perpendiculars to a line segment AB (see fig.). Show that CD bisects AB.

##### Answer:

In $\u2206$BOC and $\u2206$AOD,

∠OBC = ∠OAD

[each 90$\xb0$ (given)]

∠BOC = ∠AOD

(vertically opposite angles)

BC = AD (given)

$\therefore $ $\u2206$BOC $\cong $ $\u2206$AOD

(AAS criteria of congruency)

$\Rightarrow $ OB = OA

and OC = OD

(Congruent parts of congruent triangles)

It implies that O is the mid-point of line-segment
AB and CD.

##### Question 4:

l and m are two parallel lines intersected by another pair of parallel lines p and q (see fig.) Show that $\u2206$ABC $\cong $ $\u2206$CDA.

##### Answer:

Given than l || m

AC being a transversal.

Therefore ∠ACB = ∠DAC (alternate angles)

p || q (given)

AC being a transversal.

Therefore∠BAC = ∠ACD (alternate angles)

Now in $\u2206$ABC and $\u2206$CDA,

∠ACB = ∠DAC (Proved above)

∠BAC = ∠ACD (Proved above)

AC = AC (Common)

$\therefore $ $\u2206$ABC $\cong $ $\u2206$CDA

(AAS criteria of congruency)

##### Question 5:

Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars
from B to the arms of ∠A (see fig.). Show that:

(i) $\u2206$APB $\cong $ $\u2206$AQB

(ii) BP = BQ or B is equidistant from the
arms of ∠A.

##### Answer:

Given that line l bisects ∠A.

$\therefore $ ∠BAP = ∠BAQ

Now in $\u2206$APB and $\u2206$AQB,

∠BAP = ∠BAQ (given)

∠BPA = ∠BQA [each 90$\xb0$ (given)]

AB = AB (common)

$\therefore $ $\u2206$APB $\cong $ $\u2206$AQB

(AAS criteria of congruency)

$\Rightarrow $ BP = BQ

Congruent parts of congruent triangles i.e., B is

equidistant from the arms of ∠A.

##### Question 6:

In fig., AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

##### Answer:

Given that

∠BAD = ∠EAC

Adding ∠DAC on both sides, we get

∠BAD + ∠DAC = ∠EAC + ∠DAC

$\Rightarrow $ ∠BAC = ∠EAD ... (i)

Now in $\u2206$ABC and $\u2206$AED,

AB = AD (given)

AC = AE (given)

∠BAC = ∠EAD [From (i)]

$\therefore $ $\u2206$ABC $\cong $ $\u2206$AED

(AAS criteria of congruency)

$\Rightarrow $ BC = DE

(Congruent parts of Congruent triangles)

##### Question 7:

AB is a line segment and P is its mid-point. D and E are points on the same side of AB such
that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see fig.). Show that

(i) $\u2206$DAP $\cong $ $\u2206$EBP

(ii) AD = BE

##### Answer:

Given that

∠EPA = ∠DPB

Adding ∠EPD on both sides, we get:

∠EPA + ∠EPD = ∠DPB + ∠EPD

$\Rightarrow $ ∠APD = ∠BPE ... (i)

Now in $\u2206$DPA and $\u2206$EBP,

∠PAD = ∠PBE [$\because $ ∠BAD = ∠ABE

(given) $\therefore $ ∠PAD = ∠PBE]

∠APD = ∠BPE [From (i)]

AP = PB

[$\because $ P is the mid-point of AB (given)]

$\therefore $ $\u2206$DPA $\cong $ $\u2206$EBP

[AAS criteria of congruency]

$\Rightarrow $ AD = BE

(Congruent parts of congruent triangles)

##### Question 8:

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see fig.). Show that:

$\u2206$AMC $\cong $ $\u2206$BMD

##### Answer:

In $\u2206$AMC and $\u2206$BMD,

AM = BM

[$\because $ M is the mid-point of hyp. AB (given)]

∠AMC = ∠BMD

(vertically opp. angles)

CM = DM (given)

$\therefore $ $\u2206$AMC $\cong $ $\u2206$BMD

(SAS criteria of congruency)

$\therefore $ ∠ACM = ∠BDM ... (a)

∠CAM = ∠DBM

and AC = BD (Congruent parts of
Congruent triangles)

##### Question 9:

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see fig.). Show that:

∠DBC is a right angle.

##### Answer:

For two lines AC and DB and transversal DC,

we have

∠ACD = ∠BDC (alternate angles)

[$\because $ ∠ACM = ∠BDM using (a)

$\therefore $ ∠ACD = ∠BDC]

$\therefore $ AC || DB [$\because $ If alternate angles are equal then lines are parallel]

Now for parallel lines AC and DB and for transversal BC.

∠DBC = ∠ACB (alternate angles)

... (b)

But $\u2206$ABC is a rt. ∠d triangle, right angled at C.

$\therefore $ ∠ACB = 90$\xb0$ ... (c)

Therefore ∠DBC = 90$\xb0$ [Using (b) and (c)]

It implies that ∠DBC is a right triangle.

##### Question 10:

In right triangle ABC, right angled at C, M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see fig.). Show that:

$\u2206$DBC $\cong $ $\u2206$ACB

##### Answer:

Now in $\u2206$DBC and $\u2206$ACB,

DB = AC [Proved in part (i)]

∠DBC = ∠ACB

[each 90$\xb0$ proved in part (ii)]

BC = BC (common)

$\therefore $ $\u2206$DBC $\cong $ $\u2206$ACB

(SAS criteria of congruency)

##### Question 11:

##### Answer:

In part (iii) we have proved that

$\u2206$DBC $\cong $ $\u2206$ACB

$\therefore $ DC = AB

$\Rightarrow $DM + CM = AB

$\Rightarrow $CM + CM = AB [$\because $ DM = CM (given)]

$\Rightarrow $ 2 CM = AB

##### Question 12:

In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other
at O. Join A to O show that:

OB = OC

##### Answer:

ABC is an isosceles triangle in which AB = AC

$\therefore $ ∠C = ∠B

(angles opposite to equal sides)

$\Rightarrow $ ∠OCA + ∠OCB = ∠OBA + ∠OBC

$\Rightarrow $ ∠OCB + ∠OCB = ∠OBC + ∠OBC

$\Rightarrow $ 2∠OCB = 2∠OBC

$\Rightarrow $ ∠OCB = ∠OBC

Now in $\u2206$OBC,

∠OCB = ∠OBC (proved above)

$\therefore $ OB = OC (sides opposite to equal angles)

##### Question 13:

In an isosceles triangle ABC, with AB = AC, the bisectors of ∠B and ∠C intersect each other
at O. Join A to O show that:

AO bisects ∠A.

##### Answer:

Now in $\u2206$AOB and $\u2206$AOC

AB = AC (given)

∠OBA = ∠OCA

$\because $ ∠B = ∠C

BO bisects ∠B and CO bisects ∠C.

$\Rightarrow $ ∠OBA = ∠OCA

AO = AO (Common)

$\therefore $ $\u2206$AOB $\cong $ $\u2206$AOC

(SAS criteria of congruency)

$\Rightarrow $ ∠OAB = ∠OAC

(Congruent parts of congruent triangles)

Hence AO; bisects ∠A.

##### Question 14:

In $\u2206$ABC, AD is the perpendicular bisector of BC (see fig.). Show that $\u2206$ABC is an isosceles triangle in which AB = AC.

##### Answer:

In $\u2206$ABD and $\u2206$ACD,

BD = CD

($\because $ AD bisects BC) [given]

∠ADB = ∠ADC

= each 90$\xb0$ ($\because $ AD $\perp $ BC) [given]

AD = AD [Common]

$\therefore $ $\u2206$ABD $\cong $ $\u2206$ACD

(SAS criteria of congruency)

$\Rightarrow $ AB = AC

(Congruent parts of congruent triangles)

Therefore ABC is an isosceles triangle.

##### Question 15:

ABC is an isosceles triangle in which altitudes BE and CF are drawn to sides AC and AB respectively (see fig.). Show that these altitudes are equal.

##### Answer:

In $\u2206$ABE and $\u2206$ACF,

∠A = ∠A [Common]

∠AEB = ∠AFC (each = 90$\xb0$) [given]

AB = AC [given]

$\therefore $ $\u2206$ABE $\cong $ $\u2206$ACF

(AAS criteria of congruency)

Therefore

BE = CF

(Congruent parts of congruent triangles)

In other words, altitudes drawn to respective equal sides are equal.

##### Question 16:

ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal (see fig.). Show
that

(i) $\u2206$ABE $\cong $ $\u2206$ACF

(ii) AB = AC i.e., ABC is an isosceles triangle.

##### Answer:

In $\u2206$ABE and $\u2206$ACF,

∠A = ∠A (Common)

∠AEB = ∠AEC (each = 90$\xb0$)

[$\because $ BE $\perp $ AC and CF $\perp $AB (given)]

BE = CF (given)

(i) $\therefore $ $\u2206$ABE $\cong $ $\u2206$ACF

[AAS criteria of congruency]

(ii) So, AB = AC

[Congruent parts of congruent triangles]

i.e., $\u2206$ABC is an isosceles triangle.

##### Question 17:

ABC and DBC are two isosceles triangles on the same base BC (see fig.). Show that ∠ABD = ∠ACD.

##### Answer:

In isosceles $\u2206$ABC,

AB = AC (given)

$\therefore $ ∠ACB = ∠ABC ... (i)

(angles opposite to equal sides)

Also in isosceles $\u2206$BCD,

BD = DC

$\therefore $ ∠BCD = ∠CBD ... (ii)

(angles opposite to equal sides)

Adding corresponding sides of (i) and (ii),

∠ACB + ∠BCD = ∠ABC + ∠CBD

$\Rightarrow $ ∠ACD = ∠ABD

or ∠ABD = ∠ACD [Hence Proved]

##### Question 18:

$\u2206$ABC is an isosceles triangle in which AB = AC. Side BA is produced to D such that AD = AB (see fig.). Show that ∠BCD is a right angle.

##### Answer:

In an isosceles triangle ABC;

AB = AC

$\therefore $ ∠ACB = ∠ABC ... (i)

(angles opposite to equal sides)

Now AD = AB (By construction)

But AB = AC (given)

$\therefore $ AD = AB = AC

$\Rightarrow $ AD = AC

Now in $\u2206$ADC,

AD = AC

$\Rightarrow $ ∠ADC = ∠ACD ... (ii)

[angles opposite to equal sides of $\u2206$ADC]

From Fig., we have

∠BAC + ∠CAD = 180$\xb0$ ... (iii)

(linear pair)

As we know that exterior angle of a triangle is equal to the sum of its interior opposite
angles.

$\therefore $ In $\u2206$ABC,

∠CAD = ∠ABC + ∠ACB

= ∠ACB + ∠ACB [Using (i)]

$\Rightarrow $ ∠CAD = 2∠ACB ... (iv)

Similarly for $\u2206$ADC;

∠BAC = ∠ACD + ∠ADC

[Same reason as above]

= ∠ACD + ∠ACD[Using (ii)]

$\Rightarrow $ ∠BAC = 2∠ACD ... (v)

From (iii), (iv) and (v), we get:

2∠ACB + 2∠ACD=180$\xb0$

or 2(∠ACB + ∠ACD)=180$\xb0$

##### Question 19:

ABC is a right angled triangle in which ∠A = 90$\xb0$ and AB = AC. Find ∠B and ∠C.

##### Answer:

ABC is a right triangle in which

∠A = 90$\xb0$

and AB = AC

AB = AC

$\Rightarrow $ ∠C = ∠B ... (i)

(angles opposite to equal sides)

Now in $\u2206$ABC,

∠A + ∠B + ∠C = 180$\xb0$

(angles sum property)

$\Rightarrow $ 90$\xb0$ + ∠B + ∠B = 180$\xb0$

[$\because $ ∠A = 90$\xb0$ (Given) and ∠B = ∠C (from (i)]

$\Rightarrow $ 2∠B = 180$\xb0$ – 90$\xb0$

$\Rightarrow $ 2∠B = 90$\xb0$

$\Rightarrow $ ∠B = 45$\xb0$

Also ∠C = ∠B

$\Rightarrow $ ∠C = 45$\xb0$

##### Question 20:

Show that the angles of an equilateral triangle are 60$\xb0$ each.

##### Answer:

Let ABC be an equilateral triangle

$\therefore $ AB = BC = ACAB = BC

$\Rightarrow $ ∠C = ∠A .. (i)

(angles opp. to equal sides)

Similarly

AB = AC

$\Rightarrow $ ∠C = ∠B ... (ii)

(angles opp. to equal sides)

From (i) and (ii), we get

∠A = ∠B = ∠C ... (iii)

Now in $\u2206$ABC;

∠A + ∠B + ∠C= 180$\xb0$ ... (iv)

[angles sum property]

$\Rightarrow $ ∠A + ∠A + ∠A=180$\xb0$

$\Rightarrow $ 3∠A=180º

$\Rightarrow $ ∠A=60$\xb0$

From (iii); ∠A = ∠B = ∠C

$\Rightarrow $ ∠A = ∠B = ∠C = 60$\xb0$

Hence each angle of an equilateral triangle is 60$\xb0$.

##### Question 21:

$\u2206$ABC and $\u2206$DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see fig.). If AD is extended to intersect BC at P, show that

$\u2206$ABD $\cong $ $\u2206$ACD

##### Answer:

$\u2206$ABC is an isosceles triangle.

$\therefore $ AB = AC

$\u2206$DBC is also an isosceles triangle.

$\therefore $ BD = CD

Now in $\u2206$ABD and $\u2206$ACD

AB = AC (given)

BD = CD (given)

AD = AD (common)

$\u2206$ABD $\cong $ $\u2206$ACD

(SSS criteria of congruency)

Proved part (i)

$\Rightarrow $ ∠BAD = ∠CAD ... (a)

[Congruent parts of congruent triangles]

##### Question 22:

$\u2206$ABC and $\u2206$DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see fig.). If AD is extended to intersect BC at P, show that

$\u2206$ABP $\cong $ $\u2206$ACP

##### Answer:

$\u2206$ABC is an isosceles triangle.

$\therefore $ AB = AC

$\u2206$DBC is also an isosceles triangle.

$\therefore $ BD = CD

Now in $\u2206$ABD and $\u2206$ACD

AB = AC (given)

BD = CD (given)

AD = AD (common)

Now in $\u2206$ABP and $\u2206$ACP

AB = AC (given)

∠BAD = ∠CAD [Using (a)]

AP = AP (common)

$\therefore $ $\u2206$ABP $\cong $ $\u2206$ACP

(SAS criteria of congruency)

##### Question 23:

$\u2206$ABC and $\u2206$DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see fig.). If AD is extended to intersect BC at P, show that

AP bisects ∠A as well as ∠D.

##### Answer:

$\u2206$ABC is an isosceles triangle.

$\therefore $ AB = AC

$\u2206$DBC is also an isosceles triangle.

$\therefore $ BD = CD

Now in $\u2206$ABD and $\u2206$ACD

AB = AC (given)

BD = CD (given)

AD = AD (common)

In part (ii), we have proved that

$\u2206$ABP $\cong $ $\u2206$ACP

So, ∠BAP = ∠CAP

(Congruent parts of congruent triangles)

It implies that AP bisects ∠A.

In part (i) we have proved that

$\u2206$ABD $\cong $ $\u2206$ACD

So, ∠ADB = ∠ADC ...(b)

(Congruent parts of congruent triangles)

∠ADB + ∠BDP = 180$\xb0$

(linear pair) ... (c)

∠ADC + ∠CDP = 180$\xb0$ ... (d)

(linear pair)

From (c) and (d), we get:

∠ADB + ∠BDP = ∠ADC + ∠CDP

or ∠ADB + ∠BDP = ∠ADB + ∠CDP

[Using (b)]

$\Rightarrow $ ∠BDP = ∠CDP

$\Rightarrow $ DP bisects ∠D.

or we can say that AP bisects ∠D.

##### Question 24:

AP is the perpendicular bisector of BC.

##### Answer:

$\u2206$ABC is an isosceles triangle.

$\therefore $ AB = AC

$\u2206$DBC is also an isosceles triangle.

$\therefore $ BD = CD

Now in $\u2206$ABD and $\u2206$ACD

AB = AC (given)

BD = CD (given)

AD = AD (common)

In part (ii) we have proved that

$\u2206$ABP $\cong $ $\u2206$ACP

$\therefore $ BP = PC ...(e)

(Congruent parts of congruent triangles)

and ∠APB = ∠APC (c.p.c.t.) ... (f)

Now ∠APB + ∠APC = 180$\xb0$ (linear pair)

$\Rightarrow $ ∠APB + ∠APB = 180$\xb0$ [Using (f)]

$\Rightarrow $ 2∠APB = 180$\xb0$

$\Rightarrow $ ∠APB = 90$\xb0$

$\Rightarrow $ AP $\perp $ BC ... (g)

From (e) we have BP = PC and from (g), we have

proved AP $\perp $ BC. So, collectively we can say that

AP is perpendicular bisector of BC.

##### Question 25:

AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

(i) AD bisects BC

(ii) AD bisects ∠A.

##### Answer:

In $\u2206$ABD and $\u2206$ACD,

AB = AC (given)

∠ADB = ∠ADC (each = 90$\xb0$)

[$\because $ AD $\perp $ BC (given)]

AD = AD (common)

$\therefore $ $\u2206$ABD $\cong $ $\u2206$ACD

(RHS rule of congruency)

So, BD = DC

(Congruent parts of congruent triangles)

It implies AD; bisects BC. (Proved part (i))

Also ∠BAD = ∠CAD

(Congruent parts of congruent triangles)

It implies AD bisects ∠A.

##### Question 26:

Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of $\u2206$PQR (see fig). Show that:

$\u2206$ABM $\cong $ $\u2206$PQN

##### Answer:

Now in $\u2206$ABM and $\u2206$PQN,

AB = PQ (given)

AM = PN (given)

BM = QN [proved in part (c)]

$\therefore $ $\u2206$ABM $\cong $ $\u2206$PQN

(SSS criteria of congruency)

So, ∠B = ∠Q ... (d)

(Congruent parts of congruent triangles)

##### Question 27:

Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of $\u2206$PQR (see fig). Show that:

$\u2206$ABC $\cong $ $\u2206$PQR

##### Answer:

In $\u2206$ABC and $\u2206$PQR,

AB = PQ (given)

∠B = ∠Q [Using part (d)]

BC = QR (given)

$\therefore $ $\u2206$ABC $\cong $ $\u2206$PQR

(SAS criteria of congruency)

##### Question 28:

BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.

##### Answer:

In $\u2206$BEC and $\u2206$CFB,

∠BEC = ∠CFB (each = 90$\xb0$)

[$\because $ BE $\perp $ AC and CF $\perp $ AB]

BC = BC (common)

BE = CF (given)

$\therefore $ $\u2206$BEC $\cong $ $\u2206$CFB
(RHS rule of congruency)

So, EC = FB ...(i)

(Congruent parts of congruent triangles)

Now, in $\u2206$AEB and $\u2206$AFC,

∠A = ∠A (common)

∠AEB = ∠AFC

(each = 90$\xb0$) [given]

EB = FC (given)

$\therefore $ $\u2206$AEB $\cong $ $\u2206$AFC

(AAS criteria of congruency)

So, AE = AF ...(ii)

(Congruent parts of congruent triangles)

Adding (i) and (ii), we get:

EC + AE= FB + AF

$\Rightarrow $ AC = AB

Now, in $\u2206$ABC; we have

AB = AC

It implies that $\u2206$ABC is isosceles.

##### Question 29:

ABC is an isosceles triangle with AB = AC. Draw AP $\perp $ BC and show that ∠B = ∠C.

##### Answer:

Given: ABC is an isosceles triangle in which

AB=AC

To prove: ∠B=∠C

Construction: Draw AP $\perp $ BC.

Proof: In $\u2206$ABP and $\u2206$ACP,

∠APB=∠APC

(each = 90$\xb0$) [By construction]

AB =AC (given)

AP=AP (common)

$\therefore $ $\u2206$ABP$\cong $ $\u2206$ACP

(RHS rule of congruency)

So, ∠B=∠C

(Congruent parts of congruent triangles)

##### Question 30:

Show that in a right angled triangle, the hypotenuse is the longest side.

##### Answer:

Let ABC be a right angled triangle, rt. angle at B. We have to prove that AC, the hypotenuse is the longest side.

In rt. $\u2206$ABC;

$\Rightarrow $ ∠A + ∠B + ∠C= 180$\xb0$

∠A + 90$\xb0$+ ∠C = 180$\xb0$ [Since, ∠B = 90$\xb0$]

$\Rightarrow $ ∠A + ∠C= 180$\xb0$ – 90$\xb0$ = 90$\xb0$

Now ∠A + ∠C= 90$\xb0$

and ∠B= 90$\xb0$

$\Rightarrow $ ∠B >∠C and ∠B > ∠A

As we know that the greater angle has a longer

side opposite to it.

AC > AB and AC > BC.

In other words, ∠B being the greatest angle has the longest opposite side AC i.e.,
hypotenuse.

##### Question 31:

In fig., sides AB and AC of $\u2206$ABC are extended to points P and Q respectively. Also, ∠PBC < ∠QCB. Show that AC> AB.

##### Answer:

∠ABC + ∠PBC = 108$\xb0$ ...(i)

(linear pair axiom)

∠ACB + ∠QCB = 180$\xb0$ ...(ii)

(linear pair axiom)

From (i) and (ii), we get:

∠ABC + ∠PBC = ∠ACB + ∠QCB ...(iii)

But ∠PBC, ∠QCB (given) ...(iv)

$\therefore $ Using (iv) in (iii), we have

∠ABC > ∠ACB

Now in $\u2206$ABC

∠ABC > ∠ACB

[Proved above in (iv)]

$\therefore $ AC > AB

[Since, Side opposite to greater angle is longer.]

##### Question 32:

In fig., ∠B < ∠A and ∠C < ∠D. Show that AD < BC.

##### Answer:

In $\u2206$AOB;

∠B < ∠A (given)

or ∠A > ∠B

So; OB > OA ...(i)

[Since, Side opposite to greater angle is longer]

In $\u2206$ COD;

∠C < ∠D (given)

or ∠D > ∠C

$\therefore $ OC > OD ...(ii)

[Since, Side opposite to greater angle is longer]

Adding (i) and (ii), we get:

OB + OC > OA + OD

$\Rightarrow $ BC > AD

or AD < BC [Hence Proved]

##### Question 33:

AB and CD are respectively the smallest and longest sides of a quadrilateral ABCD (see fig., ). Show that ∠A >∠C and ∠B > ∠D.

##### Answer:

Given: ABCD is a quadrilateral. AB is the shortest side and CD is the longest side.

To prove: ∠A > ∠C and ∠B > ∠D.

Construction: Join A and C also B and D.

Proof: In $\u2206$ABC, AB is the smallest side

$\therefore $ BC > AB

$\Rightarrow $ ∠1 > ∠2 ...(i)

[Since, Angle opposite to longer side is greater]

In $\u2206$ADC, CD is the largest side

$\therefore $ CD > AD

$\Rightarrow $ ∠3 > ∠4 ...(ii)

Adding (i) and (ii), we get

∠1 + ∠3 > ∠2 + ∠4

$\Rightarrow $ ∠A >∠C [This proves 1st part]

Now in $\u2206$ADB, AB is the smallest side

$\therefore $ AD > AB

$\Rightarrow $ ∠5 > ∠6 ...(iii)

and in $\u2206$BCD, CD is the greatest side

$\therefore $ CD > BC $\Rightarrow $ ∠7 > ∠8 ...(iv)

Adding (iii) and (iv), we get:

∠5 + ∠7 > ∠6 + ∠8

∠B > ∠D [This proves 2nd part]

Hence ∠A > ∠C and ∠B > ∠D

[Hence Proved]

##### Question 34:

In fig. PR >PQ and PS bisects ∠QPR. Prove that ∠PSR > ∠PSQ.

##### Answer:

In $\u2206$PQR,

PR > PQ (given)

$\therefore $ ∠PQR > ∠PRQ

(∠opp. to longer side is greater) ...(1)

Again ∠1 = ∠2

[Since, PS is the bisector of ∠P] ...(2)

$\therefore $ ∠PQR + ∠1 > ∠PRQ + ∠2 ...(3)

But ∠PQS + ∠1 + ∠PSQ =

∠PRS + ∠2 + ∠PSR = 180$\xb0$

[Sum of ∠s of $\u2206$ is 180$\xb0$]

$\therefore $ ∠PQR + ∠1 + ∠PSQ =

∠PRQ + ∠2 + ∠PSR ...(4)

[Since ∠PRS = ∠PRQ and

∠PQS = ∠PQR]

From (3) and (4), we get

∠PSQ < ∠PSR

or ∠PSR > ∠PSQ

##### Question 35:

Show that of all line segments drawn from a given point not on it, the perpendicular line segment is the shortest.

##### Answer:

Given: l is a line and P is point not lying on l.

PM $\perp $ l.

N is any point on l other than M.

To prove: PM < PN

Proof: In $\u2206$PMN, ∠M is the right angle. So, N is an acute angle.

(Angle sum property of triangle)

$\therefore $ ∠M > ∠N.

$\therefore $ PN > PM (Side opp. greater angle)

or PM < PN.

Hence, of all line segments drawn from a given point not on it, the perpendicular is
the shortest.

##### Question 36:

ABC is a triangle. Locate a point in the interior of $\u2206$ ABC which is equidistant from all the vertices of $\u2206$ABC.

##### Answer:

Let ABC be a triangle.

Draw perpendicular bisectors PQ and RS of sides AB and BC respectively of triangle ABC.

Let PQ bisects AB at M and RS bisects BC at point N.

Let PQ and RS intersect at point O.

Join OA, OB and OC.

Now in $\u2206$AOM and BOM

AM = MB (By construction)

∠AMO = ∠BMO

(each = 90$\xb0$) [By construction]

OM = OM (Common)

$\therefore $ $\u2206$AOM $\cong $ $\u2206$BOM

(SAS criteria of congruency)

So, OA = OB (c.p.c.t.) ....(i)

Similarly, $\u2206$BON $\cong $ $\u2206$CON

$\Rightarrow $ OB = OC (c.p.c.t) ....(ii)

From (i) and (ii), we observe that

OA = OB = OC

Hence, O is the point of intersection of perpendicular bisectors of any two sides of $\u2206$ABC equidistant from its vertices.

##### Question 37:

In a triangle locate a point in its interior which is equidistant from all the sides of the triangle.

##### Answer:

Let ABC be a triangle.

Draw bisectors of ∠B and ∠C.

Let these angle bisectors intersect each other at

point I.

Draw IK $\perp $ BC

Also draw IJ $\perp $ AB

and IL $\perp $ AC

Join AI.

In $\u2206$BIK and $\u2206$BIJ,

∠IKB = ∠IJB (each = 90$\xb0$)

(By construction)

∠IBK = ∠IBJ

[$\because $ BI is the bisector of ∠B]

(By construction)

BI = BI (Common)

$\therefore $ $\u2206$BIK $\cong $ $\u2206$BIJ

(AAS criteria of congruency)

$\therefore $ IK = IJ

(Congruent parts of congruent triangles) ....(i)

Similarly, $\u2206$CIK $\cong $ $\u2206$CIL

So IK = IL (c.p.c.t) ....(ii)

From (i) and (ii), we get:

IJ = IK = IL ;

Hence, I is the point of intersection of angle bisectors of any two angles of $\u2206$ABC equidistant from its sides.

##### Question 38:

In a huge park, people are concentrated at three points (see fig.).

A: where there are different slides and swings for children.

B: near which a man-made lake is situated.

C: which is near to a large parking and exit.

Where should an icecream parlour be set up so that maximum number of persons can approach
it?

##### Answer:

The parlour should be equidistant from A, B

and C.

For this let we draw perpendicular bisector say l of line joining points B and C also draw
perpendicular bisector say m of line joining points A and C.

Let l and m intersect each other at point O.

Now point O is equidistant from points A, B and C.

Join OA, OB and OC.

Proof: In $\u2206$BOP and $\u2206$COP

OP = OP (Common)

∠OPB = ∠OPC (each = 90$\xb0$)

(By Const.)

BP = PC

[$\because $ P is the mid-point of BC]

$\therefore $ $\u2206$BOP $\cong $ $\u2206$COP

[SAS criteria of congruency]

So OB = OC (c.p.c.t.) ...(i)

Similarly $\u2206$AOQ $\cong $ $\u2206$COQ

$\Rightarrow $ OA = OC (c.p.c.t.) ..(ii)

From (i) and (ii), we get:

OA = OB = OC

We observe that ice-cream parlour should be set up at point O; the point of intersection of
perpendicular bisectors of any two sides out of three formed by joining these points.

##### Question 39:

Complete the hexagonal and star shaped Rangolies (see figs.) by filling them with as many equilateral triangles of side 1 cm as you can. Count the number of triangles in each case. Which has more triangles?

##### Answer:

In hexagonal rangoli: Number of equilateral triangles each of side 5 cm = 6

$\therefore $ Area of equilateral triangle of side 5 cm

Area of hexagonal rangoli = 6 × area of an equilateral triangle

Now in Star Rangoli

Number of equilateral triangles each of side 5 cm = 12

So, total area of star rangoli = 12 × area of an equilateral triangle of side 5 cm

From (iii) and (v), we observe that star rangoli has more equilateral triangles each of side 1 cm.