# Chapter 7: Triangles

Exercise 7.1

Question 1:

In quadrilateral PQRS, PR = PQ and PQ bisects $$\angle P$$ (look at the fig.). Show that the $$\triangle PQR\ =\ \triangle PQS$$. What can you say about QR and QS?

Solution:

In $$\triangle PQR$$ and $$\triangle PQS$$ , we have

PR = PS

$$\angle RPQ = \angle SPQ$$ (PQ bisects $$\angle P$$)

PQ = PQ (common)

$$\triangle PQR = \triangle PQS$$ (By SAS congruence)

Hence Proved.

Therefore, QR = QS (CPCT)

Question 2:

ABCD is a quadrilateral in which AD = BC and $$\angle DAB$$= $$\angle CBA$$ (see Fig.). Prove that

(i) $$\triangle ABD = \triangle BAC$$

(ii) BD = AC

(iii) $$\angle ABD = \angle BAC$$

Solution:

In the given figure, ABCD is a quadrilateral in which AD = BC and $$\angle DAB$$= $$\angle CBA$$.

In $$\triangle DAB\ and\ \triangle BAC$$, we have

$$\angle DAB = \angle CBA$$ [Given]

AB = AB [Common]

$$\triangle ABD = \triangle BAC$$ [By SAS congruence]

BD = AC [CPCT]

and $$\angle ABD = \angle BAC$$ [CPCT]

Hence Proved.

Question 3:

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

Solution:

In $$\triangle AOD = \triangle BOC$$, we have,

$$\angle AOD = \angle BOC$$ [Vertically opposite angles)

$$\angle CBO = \angle DAO (Each = 90^{o})$$

$$\triangle AOD = \triangle BOC$$ [By AAS congruence]

Also, AO = BO [CPCT]

Hence, CD bisects AB Proved.

Question 4:

L and m are two para;;e; lines intersected by another pair of parallel lines p and q (see fig.). Show that $$\triangle ABC + \triangle CDA$$ .

Solution:

In the given figure, ABCD is a parallelogram in which AC is a diagonal, i.e., AB || DC and BC || AD.

In $$\triangle ABC \, and\, \triangle CDA$$, we have,

$$\angle BAC = \angle DCA$$ (Alternate angles)

$$\angle BCA = \angle DAC$$ (Alternate angles)

AC = AC (Common)

$$\triangle ABC = \triangle CDA$$ (By SAS congruence)

Hence Proved.

Question 5:

Line l is the bisector of an angle A and B is any point on l. BP and BQ are perpendicular from B to the arms of $$\angle A$$ (See fig.). Show that:

(i)  $$\triangle APB\ and\ \triangle AQB$$

(ii) BP = BQ or B is equidistant from the arms of $$\angle A$$

Solution:

In $$\triangle APB\ and\ \triangle AQB$$ , we have

$$\angle PAB = \angle AQB$$ (l is the bisector of $$\angle A$$)

$$\angle APB = \angle AQB\ (Each = 90^{o})$$

AB = AB

$$\triangle APB = \triangle AQB$$ (By AAS congruence)

Also, BP = BQ (By CPCT)

i.e, B is equidistant from the arms of $$\angle A$$

Hence proved.

Question 6:

In the fig, AC = AE, AB = AD and $$\triangle BAD = \triangle EAC$$ . Show that BC = DE.

Solution:

$$\angle BAD = \triangle EAC$$ (Given)

$$\angle BAD + \angle DAC = \angle EAC + \angle DAC (Adding \angle DAC\ to\ both\ sides)$$ $$\angle BAC = \angle EAC$$

Now, in $$\triangle ABC\, and\, \triangle ADE$$ , We have

AC = AE

$$\angle BAC = \angle DAE$$

$$\triangle ABC= \triangle ADE$$ (By SAS congruence)

BC = DE (CPCT)

Hence Proved.

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 $$\angle BAD = \angle ABE\ and\ \angle EPA = \angle DPB$$ (see fig.). Show that

(i) $$\triangle DAP = \triangle EBP$$

Solution:

In $$\triangle DAP\ and\ \triangle EBP$$, we have

AP = BP (P is the midpoint of the line segment AB)

$$\angle BAD = \angle ABE$$ (Given)

$$\angle EPB = \angle DPA$$ $$(\angle EPA = \angle DPB \Rightarrow \angle EPA + \angle DPE = \angle DPB + \angle DPE)$$ $$\triangle DPA = \triangle EPB$$

Hence Proved.

Question 8:

In right triangle ABC, right angles 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 :

(i) $$\triangle AMC = \triangle BMD$$

(II) $$\angle DBC$$ is a right angle.

(iii) $$\triangle DBC = \triangle ACB$$

(iv) $$CM= \frac{1}{2} AB$$

Solution:

In $$\triangle BMB and \triangle DMC$$ , we have

(i) DM = CM (given)

BM = AM (M is the midpoint of AB)

$$\angle DMB = \angle AMC$$ (Vertically opposite angles)

$$\triangle AMC = \triangle BMD$$ (By SAS)

Hence Proved.

(ii) AC || BD $$( \angle DBM\ and\ \angle CAM$$ are .alternate angles)

$$\Rightarrow \angle DBC + \angle ACB = 180^{o}$$ (Sum of co-interior angles)

$$\Rightarrow \angle DBC = 90^{o}$$

Hence proved.

(iii) In $$\triangle DBC\ amd\ \triangle ACB$$, We have

DB = AC (CPCT)

BC = BC (Common)

$$\angle DBC = \angle ACB (Each = 90^{o})$$

$$\triangle DBC = \triangle ACB$$ (By SAS)

Hence proved.

(iv) AB = CD

$$\Rightarrow \frac{1}{2} AB = \frac{1}{2} CD$$ (CPCT)

Hence, $$\frac{1}{2} AB = CM$$ Proved.

Exercise 7.2:

Question 1:

In an isosceles triangle XYZ with XY=XZ ,the bisector of $$\angle B$$ and $$\angle C$$

Intersect each other at O .join A at O .Show that:

(1)OY=OZ    (2)XO bisects $$\angle A$$

Solution:

$$\frac{1}{2}\angle XYZ=\frac{1}{2}\angle XZY$$ $$\angle ZYO=\angle XZY$$

[OY and OZ are bisector of

$$\angle Y$$ and $$\angle Z$$  respectively]

OBY=OZ [ sides opposite to equal angles are equal]

Again,     $$\frac{1}{2}\angle XYZ=\frac{1}{2}\angle XZY$$

$$\angle XYO=\angle XYO$$               [ $$∴$$OY and OZ are bisector  of $$\angle Y$$ and  $$\angle Z$$ respectively   ]

In $$\Delta XYO =\Delta XYO$$ , we have

XY=XZ  [Given]

OY=OZ  [proved XYove]

$$\angle XYO=\angle XZO$$         [proved above]

$$\Delta XYO =\Delta XZO$$          [SAS congruence]

$$\Delta XYO$$= $$\angle ZXO$$   [CPCT]

XO bisects $$\angle X$$             proved

Question 2:

In $$\Delta XYZ$$   ,XO is the perpendicular bisector of YZ .Prove that $$\Delta XYZ$$ is an  isosceles triangle in which  XY=XZ

Solution:

In $$\Delta XYO= \Delta XZO$$ , we have

$$\angle XOZ =\angle XOZ$$    [] each =90°]

YO=ZO [ XO bisects YZ]

XO=XO [common]

$$∴ \Delta XYO=\Delta XZO$$   [SAS]

$$∴$$ XY=XZ    [CPCT]

Hence, $$\Delta XYZ$$ is an isosceles triangle .Proved

Question 3:

ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and

AB respectively (see figure).Show that these altitudes are equal.

Solution:

In $$\Delta ABC,$$

AB=AC   [Given]

$$\angle B=\angle C$$     [angles opposite to equal sides of a triangle are equal]

Now in right triangles BFC and CEB ,

$$\angle BFC=\angle CEB$$                  [Each =90° ]

$$\angle FBC=\angle ECB$$                 [proved above]

BC=BC

$$∴ \Delta BFC =\Delta CEB$$            [AAS]

Hence, BE=CF [CPCT] proved

Question 4:

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

(1) $$\Delta ABE \cong \Delta ACF$$

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

Solution(1) in $$\Delta ABE$$ and ACF , we have

BE=CF         [Given]

$$\angle BAE=\angle CAF$$  [common]

$$\angle BEA=\angle CFA$$    [ Each =90°]

So, $$\Delta ABE = \angle ACF$$        [AAS]proved

(2)  also, AB=AC                 [ CPCT]

i,e., ABC is an isosceles triangle Proved.

Question 5:

ABC and DBC are two isosceles triangle on the same base BC .show that

$$\angle ABD=\angle ACD$$

Solution. In isosceles $$\Delta ABC$$, We have

AB= AC

$$\angle ABC=\angle ACB$$   ———-(i)

[Angles opposite to equal sides are equal]

Now , in isosceles $$\Delta DCB$$, We have

BD=CD

$$\angle DBC=\angle DCB$$ —————–(ii)

[Angle opposite to equal sides are equal]

Adding (i) and (ii)  , we have

$$\angle ABC+\angle DBC=\angle ACB+\angle DCB$$

$$\angle ABD=\angle ACD$$ .proved

Question 6:

$$\Delta ABC$$ is an isosceles triangle in which AB=AC side BA is produced to D such that AD =AB  .show that  $$\angle BCD$$ is a right angle.

Solution:

AB=AC

$$\angle ACB= \angle ABC$$—————-(1)

[Angles opposite to equal sides are equal]

$$∴ \angle ACD=\angle ADC$$ ——————(ii)

[Angles opposite to equal sides are equal]

$$\angle ACB+\angle ACD=\angle ABC+\angle ADC$$ $$\angle BCD=\angle ABC+\angle ADC$$

Now in $$\Delta BCD ,We have$$

$$\angle BCD+\angle DBC+\angle BDC=$$= 180  [Angle sum property of a triangle]

$$∴ \angle BCD+\angle BCD$$=180°

$$2\angle BCD=180$$ $$2\angle BCD=180$$

Question 7:

ABC is a right angled triangle in which $$\angle A$$=90 and AB=AC.Find

$$\angle B and \angle C$$

Solution.  In $$\Delta ABC$$, We have

$$\angle A$$=90

AB =AC

We know that angles opposite to equal sides of an isosceles triangle are equal

So, $$\angle B=\angle C$$

Since $$\angle A$$=90 , therefore sum of remaining two angles =90

Hence $$\angle B=\angle C=45$$

Question 8:

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

Solution:

As $$\Delta ABC$$  is an equilateral

So, AB=BC=AC

Now, AB=AC

$$\angle ACB=\angle ABC$$——-(ii)[ angles sum property of a triangle]

Again BC=AC

$$\angle BAC=\angle ABC$$—-(ii)    [same reason]

Now in $$\Delta ABC$$

$$\angle ABC+\angle ACB+\angle BAC=180$$  [angle sum property of a triangle are equal]

$$\angle ABC+\angle ACB+\angle BAC=180$$=180 [from (i) (ii)]

$$3\angle ABC$$=180

$$\angle ABC$$ =180/3=60°

Also from (i) and (ii)

$$\angle ACB$$=60 and $$\angle BAC$$

Hence each angle of an equilateral triangle is 60°  Proved

Exercise 7.3

Question 1:

On the same base BC XYZ and DYZ are two isosceles triangles and vertices X and D are on the same side of YZ. If XD is extended to intersect YZ at P, show that

(i) XYD XZD

(ii) XYP XZP

(iii) XP bisects X as well as D.

(iv) XP is the perpendicular and bisector of YZ.

Solution:

(i) In ⧍XYD and ⧍XZD, we have

XY = XZ [Given]

YD = ZD [Given]

XD = XD [Common]

⧍XYD ⩬ ⧍XZD [SSS congruence].

(ii) In ⧍XYP and ⧍XZP, we have

XY = XZ

∠YXP = ∠ZXP

XP = XP

⧍XYP ⩬ ⧍XZP [SAS congruence].

(iii)⧍XYD ⩬ ⧍XDY

∠XDY = ∠XDZ

=> 180° — ∠XDY = 180° — ∠XDZ
Also, from part (ii), ∠YXPD = ∠ZXP

XP bisects DX as well as ∠D.

(iv) Now, YP = ZP and ∠YPX = ∠ZPX

But ∠YPX + ∠ZPX = 180°

So, 2∠BPA = 180°

Or, ∠YPX = 90°

Since YP = ZP, therefore XP is perpendicular bisector of YZ.

Question 2:

Isosceles triangle XYZ has an altitude XD, in which XY = XZ. Show that

(i) XD bisects YZ

(ii) XD bisects X

Solution:

XY = XZ

XD = XD

⧍XYD ⩬ ⧍XZD.

YD = ZD

Hence, XD bisects YC.

(ii) Also, ∠YXD = ∠ZXD

Hence XD bisects ∠X.

Question 3:

Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of PQR. Show that:

(i) ABM   APQN (ii) ABC PQR

Solution:

(i) In ⧍ABM and ⧍PQN,

we have

BM = QN

$$\frac{1}{2}BC = \frac{1}{2}QR$$

AB = PQ

AM = PN

⧍ABM = ⧍PQN [ SSS Congruence ]

∠ABM = ∠PQN.

(ii) Now, in ⧍ABC and ⧍PQR, we have

AB = PQ

∠ABC = ∠PQR

BC = QR

⧍ABC ⩬  ⧍PQR [SAS congruence]

Question 4:

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

Solution:

BE and CF are altitudes of a ⧍ABC.

∠BEC = ∠CFB = 90°

Now, in right triangles BEB and CFB, we have

Hyp. BC = Hyp. BC

Side BE= Side CF

⧍BEC ⩬  ⧍CFB

∠BCE = ∠CBF

Now, in ⧍ABC, ∠B = ∠C

AB = AC

Hence, ⧍ABC is an isosceles triangle.

Question 5:

AB = AC in an isosceles triangle ABC. Draw AP perpendicular BC to show that ∠B= ∠C.

Solution:

Draw AP perpendicular BC. In ⧍ABP and ⧍ACP, we have

AB = AC

∠APB = ∠APC

AB = AP [Common]

⧍ABP = ⧍ACP [By RHS congruence rule]

Also, ∠B = ∠C Proved. [CPCT]

Exercise 7.4

Question 1:

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

Solution:

ABC is a right angle triangle, right angled at B.

Now $$\angle A + \angle C = 90^{o}$$

Angles A and C are less than 90o

Now,    $$\angle B > \angle A$$

AC > BC ….(i)

(Side opposite to greater angle is longer)

Again, $$\angle B > \angle C$$

AC > AB …(ii)

(Side opposite to greater angle are longer)

Hence, from (i) and (ii), we can say that AC (hypotenuse) is the longest side.

Hence proved.

Question 2:

In the figure, sides AB and AC of $$\triangle ABC$$ are extended to points P and Q  respectively. Also, $$\angle PBC < \angle QCB$$ . Show that AC > AB.

Solution:

$$\angle ABC + \angle PBC = 180^{o} (Linear pair)$$ $$\Rightarrow \angle ABC = 180^{o} – \angle PBC$$

Similarly, $$\angle ACB = 180^{o} – \angle QCB$$

It is given that $$\angle PBC < \angle QCB$$

$$180^{o} – \angle QCB < 180^{o} – \angle PBC$$

Or $$\angle ACB < \angle ABC$$ [From (i) and (ii)]

$$\Rightarrow AB < AC$$ $$\Rightarrow AC > AB$$

Hence proved.

Question 3:

In the figure, $$\angle B < \angle A\ and\ \angle C < \angle D$$. Show that AD < BC.

Solution:

$$\angle B < \angle A$$ (Given)

BO > AO …(i)

(Side opposite to greater angle is longer)

$$\angle C < \angle D$$ (Given)

CO > DO …(ii)

(Same reason)

BO + CO > AO + DO

Hence Proved.

Question 4:

AB and CD are respectively the smallest and longest side of a quadrilateral ABCD (see fig.) Show that $$\angle A > \angle C\ and \angle B > \angle D$$ .

Solution:

Join AC.

Mark the angles as shown in the figure..

In $$\triangle ABC$$ ,

BC > AB (AB is the shortest side)

$$\angle 2 > \angle 4$$ …(i)

[Angle opposite to longer side is greater]

In $$\triangle ADC$$ ,

CD > AD (CD is the longest side)

$$\angle 1 > \angle 3$$ …(ii)

[Angle opposite to longer side is greater]

Adding (i) and (ii), we have

$$\angle 2 + \angle 1 > \angle 4 + \angle 3$$ $$\Rightarrow \angle A > \angle C$$

Similarly, by joining BD, we can prove that

$$\angle B > \angle D$$

Question 5:

In the figure, PR > PQ and PS bisects $$\angle QPR$$ . Prove that $$\angle PSR > \angle PSQ$$

Solution:

PR > PQ

$$\angle PQR > \angle PRQ$$ …(i)

[Angle opposite to longer side is greater]

$$\angle QPS > \angle RPS\ (PS\ bisects\ \angle QPR)$$ …(ii)

In $$\triangle PQS$$ , $$\angle PQS + \angle QPS + \angle PSQ = 180^{o}$$

$$\Rightarrow \angle PSQ = 180^{o} – (\angle PQS + \angle QPS)$$ …(iii)

Similarly in $$\triangle PRS$$ , $$\triangle PSR = 180^{o} – (\angle PRS + \angle RPS)$$

$$\Rightarrow \angle PSR = 180^{o} – (\angle PSR + \angle QPS)$$ [from (ii) … (iv)

From (i), we know that $$\angle PQS < \angle PSR$$

So from (iii) and (iv), $$\angle PSQ < \angle PSR$$

$$\Rightarrow \angle PSR > \angle PSQ$$

Hence proved.

Question 6:

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

Solution:

We have a line I and O is the point not on I

$$OP \perp i$$

We have to prove that OP < OQ, OP < OR and OP < OS.

OP < OS

In $$\triangle OPQ$$ , $$\angle P = 90^{o}$$

Therefore, $$\angle Q$$ is an acute angle (i.e, $$\angle Q < 90^{o}$$ )

$$\angle Q < \angle P$$

Hence, OP < OQ (Side opposite to greater angle is longer)

Similarly, we can prove that OP is shorter than OR, OS, etc.

Hence proved

Exercise 7.5

Question 1:

Find a point in the interior of $$\triangle DEF$$ which is at an equal distance or equidistant from all the vertices of $$\triangle DEF$$.

Solution:

Draw perpendicular bisectors of sides DE,

EF and FD, which meets at O.

Hence, O is the required point.

Question 2:

Find a point in the interior of a triangle such that it is at equal distances from all the sides of the triangle.

Solution:

Question 3:

People are concentrated at three points in a park namely A,B and C. (see Fig.).

A: is where swings and slides for children are present

B: is where a lake is present

C: is where there is a large parking lot and exit

Where do you think an ice – cream parlor should be set up such that the maximum number of people can access it? Draw bisectors $$\angle A, \angle B\ and\ \angle C\ of\ \triangle ABC$$. Let these angle bisectors meet at O. O is the required point.

Solution:         Join AB, BC and CA to get a triangle ABC. Draw the perpendicular bisector of AB and BC. Let them meet at O. Then O is equidistant from A, B and C. Hence, the parlor should be set up at O so that all the other points are equidistant from it.

Question 4:

Fill the star shaped and hexagonal rangolies [see fig.(i) and (ii)] by filling them with as many equilateral triangles as you can of side 1 cm. What is the number of triangles in both the cases? Which one has the most number of triangles?

Solution: