Class 9 Mathematics
Chapter 7: Triangles
Exercise 7.1 – Stepwise Solutions
Important Congruence Rules Used
SAS: If two sides and the included angle of one triangle are equal to the corresponding two sides and included angle of another triangle, then the triangles are congruent.
RHS: In two right triangles, if hypotenuse and one corresponding side are equal, then the triangles are congruent.
CPCT: Corresponding Parts of Congruent Triangles are Equal.
Q1. In quadrilateral ACBD, AC = AD and AB bisects ∠A. Show that ΔABC ≅ ΔABD. What can you say about BC and BD?
Given:
AC = AD
AB bisects ∠A
So,
∠CAB = ∠BAD
Also, AB is common to both triangles.
Now in ΔABC and ΔABD:
AC = AD
∠CAB = ∠BAD
AB = AB
Therefore,
ΔABC ≅ ΔABD by SAS congruence rule.
By CPCT,
BC = BD
Q2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA. Prove that:
(i) ΔABD ≅ ΔBAC
(ii) BD = AC
(iii) ∠ABD = ∠BAC
Given:
AD = BC
∠DAB = ∠CBA
Also, AB is common.
Now in ΔABD and ΔBAC:
AD = BC
AB = BA
∠DAB = ∠CBA
Therefore,
ΔABD ≅ ΔBAC by SAS congruence rule.
Hence by CPCT:
(ii) BD = AC
(iii) ∠ABD = ∠BAC
Q3. AD and BC are equal perpendiculars to a line segment AB. Show that CD bisects AB.
Let CD intersect AB at O.
Given:
AD ⟂ AB and BC ⟂ AB
AD = BC
So, ∠DAO = ∠CBO = 90°
Also, ∠DOA = ∠COB since they are vertically opposite angles.
Now in ΔDAO and ΔCBO:
∠DAO = ∠CBO
∠DOA = ∠COB
AD = BC
Therefore,
ΔDAO ≅ ΔCBO by AAS congruence rule.
By CPCT,
AO = OB
Hence, O is the midpoint of AB.
So, CD bisects AB.
Q4. l and m are two parallel lines intersected by another pair of parallel lines p and q. Show that ΔABC ≅ ΔCDA.
Given:
l ∥ m and p ∥ q
In the figure, AC is common to both triangles.
Since p ∥ q and AC is a transversal,
∠BAC = ∠DCA
Since l ∥ m and AC is a transversal,
∠BCA = ∠CAD
Also, AC = AC
Therefore, in ΔABC and ΔCDA:
∠BAC = ∠DCA
∠BCA = ∠CAD
AC = AC
Hence,
ΔABC ≅ ΔCDA by ASA congruence rule.
Q5. Line l is the bisector of ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A. Show that:
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A
Given:
l bisects ∠A
B lies on l
BP ⟂ one arm of ∠A
BQ ⟂ other arm of ∠A
So, ∠APB = ∠AQB = 90°
Since AB lies on the angle bisector,
∠PAB = ∠QAB
Also, AB is common in both triangles.
Now in ΔAPB and ΔAQB:
∠APB = ∠AQB
∠PAB = ∠QAB
AB = AB
Therefore,
ΔAPB ≅ ΔAQB by AAS congruence rule.
By CPCT,
BP = BQ
Hence, B is equidistant from the arms of ∠A.
Class 9 Mathematics
Chapter 7: Triangles
Exercise 7.1 – Stepwise Solutions (Questions 6 to 8)
Important Congruence Rules Used
SAS: If two sides and the included angle of one triangle are equal to the corresponding two sides and included angle of another triangle, then the triangles are congruent.
ASA: If two angles and the included side of one triangle are equal to the corresponding two angles and included side of another triangle, then the triangles are congruent.
CPCT: Corresponding Parts of Congruent Triangles are Equal.
Q6. In Fig. 7.21, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE.
Given:
AC = AE
AB = AD
∠BAD = ∠EAC
Step 1: Add ∠DAC on both sides of the given angle equality.
∠BAD + ∠DAC = ∠EAC + ∠DAC
So,
∠BAC = ∠DAE
Step 2: Consider triangles BAC and DAE.
In ΔBAC and ΔDAE:
AB = AD
AC = AE
∠BAC = ∠DAE
Therefore,
ΔBAC ≅ ΔDAE by SAS.
Step 3: By CPCT, corresponding sides are equal.
So,
BC = DE
Hence proved.
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 ∠BAD = ∠ABE and ∠EPA = ∠DPB. Show that:
(i) ΔDAP ≅ ΔEBP
(ii) AD = BE
Given:
P is the mid-point of AB
Therefore, AP = PB
Also, ∠BAD = ∠ABE
and ∠EPA = ∠DPB
Step 1: Since A, P, B are collinear,
∠DAB = ∠BAD and ∠EBA = ∠ABE are the angles at A and B respectively with line AB.
Given directly,
∠DAP = ∠EBP
Also,
∠DPA = ∠EPB
because ∠EPA = ∠DPB and they are the same angles written in reverse order.
Step 2: Consider triangles DAP and EBP.
In ΔDAP and ΔEBP:
AP = PB
∠DAP = ∠EBP
∠DPA = ∠EPB
Therefore,
ΔDAP ≅ ΔEBP by ASA.
Step 3: By CPCT,
AD = BE
Hence proved.
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) ΔAMC ≅ ΔBMD
(ii) ∠DBC is a right angle.
(iii) ΔDBC ≅ ΔACB
(iv) CM = 1/2 AB
Given:
ABC is right-angled at C
M is midpoint of AB
So, AM = MB
C, M, D are collinear and DM = CM
(i) Prove ΔAMC ≅ ΔBMD
In ΔAMC and ΔBMD:
AM = MB
CM = DM
Also, ∠AMC = ∠BMD
because these are vertically opposite angles formed by lines AB and CD intersecting at M.
Therefore,
ΔAMC ≅ ΔBMD by SAS.
(ii) Prove ∠DBC is a right angle
From part (i), by CPCT:
∠MAC = ∠MBD
Since A, M, B are collinear,
∠MAC = ∠BAC and ∠MBD = ∠ABD
Also, in right triangle ABC,
∠BAC + ∠ABC = 90°
From congruence, corresponding angle at B gives that the angle made by BD with BA equals angle BAC.
Hence, angle between DB and BC becomes 90°.
So,
∠DBC = 90°
(iii) Prove ΔDBC ≅ ΔACB
In ΔDBC and ΔACB:
∠DBC = ∠ACB = 90°
From part (i),
DB = AC
Also, BC is common.
Therefore,
ΔDBC ≅ ΔACB by SAS (or RHS, since both are right triangles).
(iv) Prove CM = 1/2 AB
Since M is the midpoint of AB,
AM = MB = 1/2 AB
From part (i),
CM = DM
and from midpoint property of hypotenuse in a right triangle,
AM = CM
Therefore,
CM = AM = 1/2 AB
Hence proved.