EXERCISE- 7.3
1.Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that:
(i) Δ ABD ≅ Δ ACD
(ii) Δ ABP ≅ Δ ACP
(iii) AP bisects ∠ A as well as ∠ D.
(iv) AP is the perpendicular bisector of BC.
Solution:
Given: Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC.
To prove:
(i) Δ ABD ≅ Δ ACD
(ii) Δ ABP ≅ Δ ACP
(iii) AP bisects ∠ A as well as ∠ D.
(iv) AP is the perpendicular bisector of BC
Proof:
In ΔABD and Δ ACD
AB = AC [given]
BD = CD [given]
AD = AD [common]
By SSS Congruence Criterion Rule
Δ ABD ≅ Δ ACD
∠ BAD = ∠CAD [CPCT]
∠ BAP = ∠CAP [CPCT] …
(ii)In ΔABP and Δ ACP
AB = AC [given]
∠ BAP = ∠CAP [proved above]
AP = AP [common]
By SAS Congruence Criterion Rule
Δ ABP ≅ Δ ACP
BP = CP [CPCT] … 2
∠APB = ∠APC [CPCT]
(iii) ∠ BAP = ∠CAP [From eq. 1]
Hence, AP bisects ∠ A.
Now, In Δ BDP and Δ CDP
BD = CD [given]
BP = CP [given]
DP = DP [common]
By SSS Congruence Criterion Rule
Δ BDP ≅ Δ CDP
∠ BDP = ∠CDP [CPCT]
AP bisects ∠ D.
(iv) AP stands on B
∠APB + ∠APC = 1800
∠APB +∠APB = 1800[proved above]
∠APB = 1800 /2
∠APB = 900
AP is the perpendicular bisector of BC.
2. AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that
(i) AD bisects BC (ii) AD bisects ∠ A.
Solution:
Given: AD is an altitude of an isosceles triangle ABC in which AB = AC.
To prove: (i) AD bisects BC
(ii) AD bisects ∠ A.
Proof: In ∆BAD and ∆CAD
∠ ADB = ∠ADC (Each 90º as AD is an altitude)
AB = AC (Given)
AD = AD (Common)
By RHS Congruence Criterion Rule
∆BAD ≅ ∆CAD
BD = CD (By CPCT)
Hence, AD bisects BC.
∠BAD = ∠CAD (By CPCT)
Hence, AD bisects ∠ A
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 Δ PQR (see Fig. 7.40). Show that:
(i) Δ ABM ≅ Δ PQN
(ii) Δ ABC ≅ Δ PQR
Solution:
Given: Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR.
To prove: (i) Δ ABM ≅ Δ PQN
(ii) Δ ABC ≅ Δ PQR
Proof: In ∆ABC, AM is the median to BC.
BM = 1/2 BC ... 1
In ∆PQR, PN is the median to QR.
QN = 1/2 QR ... 2
from eq .1 & 2
BM = QN ... 3
Now in ABM and PQN
AB = PQ (Given)
BM = QN [From equation (3)]
AM = PN [given]
By SSS congruence Criterion rule
∆ABM ≅ ∆PQN
∠B =∠Q [CPCT]
Now in∆ ABC and∆ PQR
AB = PQ [given]
∠B = ∠Q [prove above ]
BC = QR [given]
By SAS congruence Criterion rule
∆ ABC ≅ ∆ PQR
4. BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle ABC is isosceles.
Solution:
Given: BE and CF are two equal altitudes of a triangle ABC.
To prove: ABC is a isosceles.
Proof: In ∆BEC and ∆CFB,
BE = CF (Given)
∠BEC = CFB (Each 90°)
BC = CB (Common)
By RHS congruence Criterion rule
∆BEC ≅ ∆CFB
∠BCE = ∠CBF (By CPCT)
AB = AC [Sides opposite to equal angles of a triangle are equal]
Hence, ABC is isosceles.
5. ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that:
∠ B = ∠ C.
Solution:
Given: ABC is an isosceles triangle with AB = AC.
To prove: ∠ B = ∠ C.
Construction: Draw AP ⊥ BC to
Proof : In ∆APB and ∆APC
∠APB = ∠APC (Each 90º)
AB =AC (Given)
AP = AP (Common)
By RHS Congruence Criterion Rule
∆APB ≅ ∆APC
∠B = ∠C [CPCT]
