Chapter 7: Triangles
Exercise 7.1
Q1
In quadrilateral ACBD, \( AC = AD \) and AB bisects \( \angle A \) (See the given figure). Show that \( \Delta ABC \cong \Delta ABD \). What can you say about BC and BD?
Solution:
In \( \Delta ABC \) and \( \Delta ABD \):
\[ AC = AD \quad \text{(Given)} \]
\[ \angle CAB = \angle DAB \quad \text{(AB bisects } \angle A \text{)} \]
\[ AB = AB \quad \text{(Common)} \]
\[ \therefore \Delta ABC \cong \Delta ABD \quad \text{(By SAS congruence rule)} \]
\[ \Rightarrow BC = BD \quad \text{(By CPCT)} \]
Therefore, BC and BD are of equal lengths.
Q2
ABCD is a quadrilateral in which \( AD = BC \) and \( \angle DAB = \angle CBA \) (See the given figure). Prove that:
(i) \( \Delta ABD \cong \Delta BAC \)
(ii) \( BD = AC \)
(iii) \( \angle ABD = \angle BAC \)
(i) \( \Delta ABD \cong \Delta BAC \)
(ii) \( BD = AC \)
(iii) \( \angle ABD = \angle BAC \)
Solution:
In \( \Delta ABD \) and \( \Delta BAC \):
\[ AD = BC \quad \text{(Given)} \]
\[ \angle DAB = \angle CBA \quad \text{(Given)} \]
\[ AB = BA \quad \text{(Common)} \]
\[ \therefore \Delta ABD \cong \Delta BAC \quad \text{(By SAS congruence rule)} \]
\[ \Rightarrow BD = AC \quad \text{(By CPCT)} \]
\[ \text{And, } \angle ABD = \angle BAC \quad \text{(By CPCT)} \]
Q3
AD and BC are equal perpendiculars to a line segment AB (See the given figure). Show that CD bisects AB.
Solution:
In \( \Delta BOC \) and \( \Delta AOD \):
\[ \angle BOC = \angle AOD \quad \text{(Vertically opposite angles)} \]
\[ \angle CBO = \angle DAO \quad \text{(Each } 90^{\circ} \text{)} \]
\[ BC = AD \quad \text{(Given)} \]
\[ \therefore \Delta BOC \cong \Delta AOD \quad \text{(AAS congruence rule)} \]
\[ \Rightarrow BO = AO \quad \text{(By CPCT)} \]
Therefore, CD bisects AB.
Q4
\( l \) and \( m \) are two parallel lines intersected by another pair of parallel lines \( p \) and \( q \) (see the given figure). Show that \( \Delta ABC \cong \Delta CDA \).
Solution:
In \( \Delta ABC \) and \( \Delta CDA \):
\[ \angle BAC = \angle DCA \quad \text{(Alternate interior angles, as } p \parallel q \text{)} \]
\[ AC = CA \quad \text{(Common)} \]
\[ \angle BCA = \angle DAC \quad \text{(Alternate interior angles, as } l \parallel m \text{)} \]
\[ \therefore \Delta ABC \cong \Delta CDA \quad \text{(By ASA congruence rule)} \]
Q5
Line \( l \) is the bisector of an angle \( \angle A \) and B is any point on \( l \). BP and BQ are perpendiculars from B to the arms of \( \angle A \) (see the given figure). Show that:
(i) \( \Delta APB \cong \Delta AQB \)
(ii) \( BP = BQ \) or B is equidistant from the arms of \( \angle A \).
(i) \( \Delta APB \cong \Delta AQB \)
(ii) \( BP = BQ \) or B is equidistant from the arms of \( \angle A \).
Solution:
In \( \Delta APB \) and \( \Delta AQB \):
\[ \angle APB = \angle AQB \quad \text{(Each } 90^{\circ} \text{)} \]
\[ \angle PAB = \angle QAB \quad \text{(} l \text{ is the angle bisector of } \angle A \text{)} \]
\[ AB = AB \quad \text{(Common)} \]
\[ \therefore \Delta APB \cong \Delta AQB \quad \text{(By AAS congruence rule)} \]
\[ \Rightarrow BP = BQ \quad \text{(By CPCT)} \]
Or, it can be said that B is equidistant from the arms of \( \angle A \).
Q6
In the given figure, \( AC = AE \), \( AB = AD \) and \( \angle BAD = \angle EAC \). Show that \( BC = DE \).
Solution:
It is given that \( \angle BAD = \angle EAC \).
Adding \( \angle DAC \) to both sides:
\[ \angle BAD + \angle DAC = \angle EAC + \angle DAC \]
\[ \Rightarrow \angle BAC = \angle DAE \]
In \( \Delta BAC \) and \( \Delta DAE \):
\[ AB = AD \quad \text{(Given)} \]
\[ \angle BAC = \angle DAE \quad \text{(Proved above)} \]
\[ AC = AE \quad \text{(Given)} \]
\[ \therefore \Delta BAC \cong \Delta DAE \quad \text{(By SAS congruence rule)} \]
\[ \Rightarrow BC = DE \quad \text{(By CPCT)} \]
Q7
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 the given figure). Show that:
(i) \( \Delta DAP \cong \Delta EBP \)
(ii) \( AD = BE \)
(i) \( \Delta DAP \cong \Delta EBP \)
(ii) \( AD = BE \)
Solution:
It is given that \( \angle EPA = \angle DPB \).
Adding \( \angle DPE \) to both sides:
\[ \angle EPA + \angle DPE = \angle DPB + \angle DPE \]
\[ \Rightarrow \angle APD = \angle BPE \]
In \( \Delta DAP \) and \( \Delta EBP \):
\[ \angle DAP = \angle EBP \quad \text{(Given)} \]
\[ AP = BP \quad \text{(P is mid-point of AB)} \]
\[ \angle APD = \angle BPE \quad \text{(Proved above)} \]
\[ \therefore \Delta DAP \cong \Delta EBP \quad \text{(ASA congruence rule)} \]
\[ \Rightarrow AD = BE \quad \text{(By CPCT)} \]
Q8
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. Show that:
(i) \( \Delta AMC \cong \Delta BMD \)
(ii) \( \angle DBC \) is a right angle.
(iii) \( \Delta DBC \cong \Delta ACB \)
(iv) \( CM = \frac{1}{2}AB \)
(i) \( \Delta AMC \cong \Delta BMD \)
(ii) \( \angle DBC \) is a right angle.
(iii) \( \Delta DBC \cong \Delta ACB \)
(iv) \( CM = \frac{1}{2}AB \)
Solution:
(i) In \( \Delta AMC \) and \( \Delta BMD \):
\[ AM = BM \quad \text{(M is the mid-point of AB)} \]
\[ \angle AMC = \angle BMD \quad \text{(Vertically opposite angles)} \]
\[ CM = DM \quad \text{(Given)} \]
\[ \therefore \Delta AMC \cong \Delta BMD \quad \text{(By SAS congruence rule)} \]
(ii) Since \( \Delta AMC \cong \Delta BMD \):
\[ AC = BD \quad \text{(By CPCT)} \]
\[ \angle ACM = \angle BDM \quad \text{(By CPCT)} \]
Since alternate interior angles \( \angle ACM \) and \( \angle BDM \) are equal, \( DB \parallel AC \).
\[ \Rightarrow \angle DBC + \angle ACB = 180^{\circ} \quad \text{(Co-interior angles)} \]
\[ \Rightarrow \angle DBC + 90^{\circ} = 180^{\circ} \]
\[ \Rightarrow \angle DBC = 90^{\circ} \]
(iii) In \( \Delta DBC \) and \( \Delta ACB \):
\[ DB = AC \quad \text{(Proved above)} \]
\[ \angle DBC = \angle ACB \quad \text{(Each } 90^{\circ} \text{)} \]
\[ BC = CB \quad \text{(Common)} \]
\[ \therefore \Delta DBC \cong \Delta ACB \quad \text{(SAS congruence rule)} \]
(iv) Since \( \Delta DBC \cong \Delta ACB \):
\[ AB = DC \quad \text{(By CPCT)} \]
\[ \Rightarrow AB = 2 CM \quad \text{(Since } DM = CM \text{)} \]
\[ \Rightarrow CM = \frac{1}{2}AB \]
Exercise 7.2
Q1
In an isosceles triangle ABC, with \( AB = AC \), the bisectors of \( \angle B \) and \( \angle C \) intersect each other at O. Join A to O. Show that:
(i) \( OB = OC \)
(ii) AO bisects \( \angle A \)
(i) \( OB = OC \)
(ii) AO bisects \( \angle A \)
Solution:
(i) In triangle ABC, \( AB = AC \).
\[ \Rightarrow \angle ACB = \angle ABC \quad \text{(Angles opposite to equal sides are equal)} \]
\[ \Rightarrow \frac{1}{2}\angle ACB = \frac{1}{2}\angle ABC \]
\[ \Rightarrow \angle OCB = \angle OBC \]
\[ \Rightarrow OB = OC \quad \text{(Sides opposite to equal angles are equal)} \]
(ii) In \( \Delta OAB \) and \( \Delta OAC \):
\[ AO = AO \quad \text{(Common)} \]
\[ AB = AC \quad \text{(Given)} \]
\[ OB = OC \quad \text{(Proved above)} \]
\[ \therefore \Delta OAB \cong \Delta OAC \quad \text{(By SSS congruence rule)} \]
\[ \Rightarrow \angle BAO = \angle CAO \quad \text{(CPCT)} \]
Therefore, AO bisects \( \angle A \).
Q2
In \( \Delta ABC \), AD is the perpendicular bisector of BC (see the given figure). Show that \( \Delta ABC \) is an isosceles triangle in which \( AB = AC \).
Solution:
In \( \Delta ADC \) and \( \Delta ADB \):
\[ AD = AD \quad \text{(Common)} \]
\[ \angle ADC = \angle ADB \quad \text{(Each } 90^{\circ} \text{)} \]
\[ CD = BD \quad \text{(AD is the perpendicular bisector of BC)} \]
\[ \therefore \Delta ADC \cong \Delta ADB \quad \text{(By SAS congruence rule)} \]
\[ \Rightarrow AB = AC \quad \text{(By CPCT)} \]
Therefore, ABC is an isosceles triangle.
Q3
ABC is an isosceles triangle in which altitudes BE and CF are drawn to equal sides AC and AB respectively. Show that these altitudes are equal.
Solution:
In \( \Delta AEB \) and \( \Delta AFC \):
\[ \angle AEB = \angle AFC \quad \text{(Each } 90^{\circ} \text{)} \]
\[ \angle A = \angle A \quad \text{(Common angle)} \]
\[ AB = AC \quad \text{(Given)} \]
\[ \therefore \Delta AEB \cong \Delta AFC \quad \text{(By AAS congruence rule)} \]
\[ \Rightarrow BE = CF \quad \text{(By CPCT)} \]
Q4
ABC is a triangle in which altitudes BE and CF to sides AC and AB are equal. Show that:
(i) \( \Delta ABE \cong \Delta ACF \)
(ii) \( AB = AC \), i.e., ABC is an isosceles triangle.
(i) \( \Delta ABE \cong \Delta ACF \)
(ii) \( AB = AC \), i.e., ABC is an isosceles triangle.
Solution:
(i) In \( \Delta ABE \) and \( \Delta ACF \):
\[ \angle ABE = \angle ACF \quad \text{(Each } 90^{\circ} \text{)} \]
\[ \angle A = \angle A \quad \text{(Common angle)} \]
\[ BE = CF \quad \text{(Given)} \]
\[ \therefore \Delta ABE \cong \Delta ACF \quad \text{(By AAS congruence rule)} \]
(ii) Since \( \Delta ABE \cong \Delta ACF \):
\[ AB = AC \quad \text{(By CPCT)} \]
Therefore, ABC is an isosceles triangle.
Q5
ABC and DBC are two isosceles triangles on the same base BC. Show that \( \angle ABD = \angle ACD \).
Solution:
Let us join AD. In \( \Delta ABD \) and \( \Delta ACD \):
\[ AB = AC \quad \text{(Given)} \]
\[ BD = CD \quad \text{(Given)} \]
\[ AD = AD \quad \text{(Common side)} \]
\[ \therefore \Delta ABD \cong \Delta ACD \quad \text{(By SSS congruence rule)} \]
\[ \Rightarrow \angle ABD = \angle ACD \quad \text{(By CPCT)} \]
Q6
\( \Delta ABC \) 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:
In \( \Delta ABC \):
\[ AB = AC \Rightarrow \angle ACB = \angle ABC \]
In \( \Delta ACD \):
\[ AC = AD \quad \text{(Since } AD=AB \text{)} \]
\[ \Rightarrow \angle ADC = \angle ACD \]
In \( \Delta BCD \):
\[ \angle ABC + \angle BCD + \angle ADC = 180^{\circ} \]
Substituting the equal angles:
\[ \angle ACB + (\angle ACB + \angle ACD) + \angle ACD = 180^{\circ} \]
\[ 2(\angle ACB + \angle ACD) = 180^{\circ} \]
\[ 2(\angle BCD) = 180^{\circ} \]
\[ \Rightarrow \angle BCD = 90^{\circ} \]
Q7
ABC is a right angled triangle in which \( \angle A = 90^{\circ} \) and \( AB = AC \). Find \( \angle B \) and \( \angle C \).
Solution:
It is given that \( AB = AC \).
\[ \Rightarrow \angle C = \angle B \]
In \( \Delta ABC \):
\[ \angle A + \angle B + \angle C = 180^{\circ} \]
\[ 90^{\circ} + \angle B + \angle B = 180^{\circ} \]
\[ 2\angle B = 90^{\circ} \Rightarrow \angle B = 45^{\circ} \]
\[ \therefore \angle B = \angle C = 45^{\circ} \]
Q8
Show that the angles of an equilateral triangle are \( 60^{\circ} \) each.
Solution:
Let ABC be an equilateral triangle. Therefore, \( AB = BC = AC \).
\[ AB = AC \Rightarrow \angle C = \angle B \]
\[ AC = BC \Rightarrow \angle B = \angle A \]
Therefore, \( \angle A = \angle B = \angle C \).
In \( \Delta ABC \):
\[ \angle A + \angle B + \angle C = 180^{\circ} \]
\[ 3\angle A = 180^{\circ} \]
\[ \Rightarrow \angle A = 60^{\circ} \]
Hence, in an equilateral triangle, all interior angles are of measure \( 60^{\circ} \).
Exercise 7.3
Q1
\( \Delta ABC \) and \( \Delta DBC \) 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 P, show that:
(i) \( \Delta ABD \cong \Delta ACD \)
(ii) \( \Delta ABP \cong \Delta ACP \)
(iii) AP bisects \( \angle A \) as well as \( \angle D \).
(iv) AP is the perpendicular bisector of BC.
(i) \( \Delta ABD \cong \Delta ACD \)
(ii) \( \Delta ABP \cong \Delta ACP \)
(iii) AP bisects \( \angle A \) as well as \( \angle D \).
(iv) AP is the perpendicular bisector of BC.
Solution:
(i) In \( \Delta ABD \) and \( \Delta ACD \):
\[ AB = AC \quad \text{(Given)} \]
\[ BD = CD \quad \text{(Given)} \]
\[ AD = AD \quad \text{(Common)} \]
\[ \therefore \Delta ABD \cong \Delta ACD \quad \text{(By SSS congruence rule)} \]
\[ \Rightarrow \angle BAD = \angle CAD \quad \text{(By CPCT) ... (1)} \]
(ii) In \( \Delta ABP \) and \( \Delta ACP \):
\[ AB = AC \quad \text{(Given)} \]
\[ \angle BAP = \angle CAP \quad \text{[From (1)]} \]
\[ AP = AP \quad \text{(Common)} \]
\[ \therefore \Delta ABP \cong \Delta ACP \quad \text{(By SAS congruence rule)} \]
\[ \Rightarrow BP = CP \quad \text{(By CPCT) ... (2)} \]
(iii) From (1), AP bisects \( \angle A \).
In \( \Delta BDP \) and \( \Delta CDP \):
\[ BD = CD \quad \text{(Given)} \]
\[ DP = DP \quad \text{(Common)} \]
\[ BP = CP \quad \text{[From (2)]} \]
\[ \therefore \Delta BDP \cong \Delta CDP \quad \text{(By SSS rule)} \]
\[ \Rightarrow \angle BDP = \angle CDP \quad \text{(By CPCT)} \]
Hence, AP bisects \( \angle D \).
(iv) Since \( \Delta BDP \cong \Delta CDP \):
\[ \angle BPD = \angle CPD \]
\[ \angle BPD + \angle CPD = 180^{\circ} \quad \text{(Linear pair)} \]
\[ 2\angle BPD = 180^{\circ} \Rightarrow \angle BPD = 90^{\circ} \]
From (2) and this result, AP is the perpendicular bisector of BC.
Q2
AD is an altitude of an isosceles triangle ABC in which \( AB = AC \). Show that (i) AD bisects BC (ii) AD bisects \( \angle A \).
Solution:
In \( \Delta BAD \) and \( \Delta CAD \):
\[ \angle ADB = \angle ADC = 90^{\circ} \]
\[ AB = AC \quad \text{(Given)} \]
\[ AD = AD \quad \text{(Common)} \]
\[ \therefore \Delta BAD \cong \Delta CAD \quad \text{(By RHS congruence rule)} \]
(i) \( BD = CD \) (By CPCT). Hence, AD bisects BC.
(ii) \( \angle BAD = \angle CAD \) (By CPCT). Hence, AD bisects \( \angle A \).
Q3
Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of \( \Delta PQR \). Show that:
(i) \( \Delta ABM \cong \Delta PQN \)
(ii) \( \Delta ABC \cong \Delta PQR \)
(i) \( \Delta ABM \cong \Delta PQN \)
(ii) \( \Delta ABC \cong \Delta PQR \)
Solution:
(i) AM and PN are medians.
\[ BM = \frac{1}{2}BC, \quad QN = \frac{1}{2}QR \]
Since \( BC = QR \), implies \( BM = QN \).
In \( \Delta ABM \) and \( \Delta PQN \):
\[ AB = PQ \quad \text{(Given)} \]
\[ BM = QN \]
\[ AM = PN \quad \text{(Given)} \]
\[ \therefore \Delta ABM \cong \Delta PQN \quad \text{(SSS congruence rule)} \]
\[ \Rightarrow \angle B = \angle Q \quad \text{(CPCT)} \]
(ii) In \( \Delta ABC \) and \( \Delta PQR \):
\[ AB = PQ \]
\[ \angle B = \angle Q \]
\[ BC = QR \]
\[ \therefore \Delta ABC \cong \Delta PQR \quad \text{(By SAS congruence rule)} \]
Q4
BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Solution:
In \( \Delta BEC \) and \( \Delta CFB \):
\[ \angle BEC = \angle CFB = 90^{\circ} \]
\[ BC = CB \quad \text{(Common Hypotenuse)} \]
\[ BE = CF \quad \text{(Given)} \]
\[ \therefore \Delta BEC \cong \Delta CFB \quad \text{(By RHS congruence rule)} \]
\[ \Rightarrow \angle C = \angle B \quad \text{(By CPCT)} \]
\[ \therefore AB = AC \quad \text{(Sides opposite to equal angles are equal)} \]
Hence, \( \Delta ABC \) is isosceles.
Q5
ABC is an isosceles triangle with \( AB = AC \). Drawn \( AP \perp BC \) to show that \( \angle B = \angle C \).
Solution:
In \( \Delta APB \) and \( \Delta APC \):
\[ \angle APB = \angle APC = 90^{\circ} \]
\[ AB = AC \quad \text{(Given)} \]
\[ AP = AP \quad \text{(Common)} \]
\[ \therefore \Delta APB \cong \Delta APC \quad \text{(Using RHS congruence rule)} \]
\[ \Rightarrow \angle B = \angle C \quad \text{(By CPCT)} \]