Class 9 Mathematics
Chapter 7: Triangles
Exercise 7.3 – Stepwise Solutions
Important Concepts Used
1. In an isosceles triangle, equal sides subtend equal angles.
2. SAS, ASA and RHS congruence rules.
3. CPCT (Corresponding Parts of Congruent Triangles are Equal).
4. Properties of perpendicular bisector and medians.
Q1. ΔABC and ΔDBC are two isosceles triangles on the same base BC. AD is extended to meet BC at P. Show that:
(i) ΔABD ≅ ΔACD
(ii) ΔABP ≅ ΔACP
(iii) AP bisects ∠A and ∠D
(iv) AP is the perpendicular bisector of BC
Given:
In ΔABC, AB = AC
In ΔDBC, DB = DC
(i) Proof:
In ΔABD and ΔACD:
AB = AC
BD = DC
AD = AD (common)
Therefore,
ΔABD ≅ ΔACD by SSS
(ii) Proof:
From (i), ∠BAD = ∠DAC
So AD bisects ∠A
Now in ΔABP and ΔACP:
AB = AC
∠BAP = ∠PAC
AP = AP
Therefore,
ΔABP ≅ ΔACP by SAS
(iii) Result:
From congruence,
∠BAP = ∠PAC ⇒ AP bisects ∠A
Also, ∠BDA = ∠ADC ⇒ AP bisects ∠D
(iv) Result:
From (ii), BP = PC
Also, ∠APB = ∠APC = 90°
Therefore,
AP is 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 ∠A
Given: AB = AC and AD ⟂ BC
(i) Proof:
In ΔABD and ΔACD:
AB = AC
AD = AD
∠ADB = ∠ADC = 90°
Therefore,
ΔABD ≅ ΔACD by RHS
So, BD = DC
Hence AD bisects BC
(ii) Proof:
From congruence,
∠BAD = ∠CAD
Hence AD bisects ∠A
Q3. AB = PQ, BC = QR and median AM = PN. Show that:
(i) ΔABM ≅ ΔPQN
(ii) ΔABC ≅ ΔPQR
Given:
AB = PQ
BC = QR
AM = PN
Since AM and PN are medians:
BM = MC and QN = NR
(i) Proof:
In ΔABM and ΔPQN:
AB = PQ
AM = PN
BM = QN
Therefore,
ΔABM ≅ ΔPQN by SSS
(ii) Proof:
From (i), ∠ABM = ∠PQN
Now in ΔABC and ΔPQR:
AB = PQ
BC = QR
∠ABC = ∠PQR
Therefore,
ΔABC ≅ ΔPQR by SAS
Q4. BE and CF are equal altitudes of triangle ABC. Using RHS, prove that triangle ABC is isosceles.
Given: BE = CF
BE ⟂ AC and CF ⟂ AB
In right triangles ΔBEC and ΔCFB:
BE = CF
BC = BC (common)
Right angles at E and F
Therefore,
ΔBEC ≅ ΔCFB by RHS
By CPCT,
AB = AC
Hence triangle ABC is isosceles
Q5. ABC is an isosceles triangle with AB = AC. Draw AP ⟂ BC to show that ∠B = ∠C.
Construction: Draw AP ⟂ BC
Proof:
In ΔABP and ΔACP:
AB = AC
AP = AP
∠APB = ∠APC = 90°
Therefore,
ΔABP ≅ ΔACP by RHS
By CPCT,
∠ABP = ∠ACP
Thus,
∠B = ∠C