**In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see Fig. 7.16). Show that Δ ABC ≅ Δ ABD. What can you say about BC and BD?**Solution:

In ∆ABC and ∆ABD, we have AC = AD

∠CAB = ∠DAB *[line AB intersects ∠CAB]*

AB is common line segment

⸫ ∆ABC = ∆ABD, *[By SAS congruence]*

⸫ BC = AB *©*

**ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see Fig. 7.17). Prove that**

**(i) Δ ABD ≅ Δ BAC**

**(ii) BD = AC**

**(iii) ∠ ABD = ∠ BAC.**

Solution:

In the given figure, ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA

In Δ ABD = Δ BAC, we have,

AD = BC (given)

∠DAB = ∠CBA (given)

AB = AB (common line segment)

⸫ Δ ABD ≅ Δ BAC (By SAS congruence)

⸫ BD = AC (Corresponding parts of congruent triangles are congruent)

and ∠ ABD = ∠ BAC (Corresponding parts of congruent triangles are congruent)

**AD and BC are equal perpendiculars to a line segment AB (see Fig. 7.18). Show that CD bisects AB.**

Solution:

In ΔAOD and ΔBOC, we have,

∠AOD = ∠BOC *[vertically opposite angles]*

∠CBO = ∠DAO *[given]*

and AD = BC *[given]*

⸫ΔAOD ≅ ΔBOC *[By AAS congruence]*

Also, AO = BO (Corresponding parts of congruent triangles are congruent)

Hence CD bisects AB

**l and m are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that Δ ABC ≅ Δ CDA.**

Solution:

Given l || m and p||q

To prove ΔABC ≅ ΔCDA

In ΔABC and ΔCDA

∠BCA = ∠DAC (alternate interior angles)

AC = CA ( common)

∠BAC = ∠DCA (alternate interior angles)

⸫ ΔABC ≅ ΔCDA (ASA congruence)

**line l is the bisector of an angle ∠ A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠ A (see Fig. 7.20). Show that:**

**(i) Δ APB ≅ Δ AQB**

**(ii) BP = BQ or B is equidistant from the arms of ∠ A.**

Solution:

I is the bisector of an ∠ A

BP and BQ are perpendiculars.

(i) In Δ APB and Δ AQB

∠P = ∠Q (right angles)

∠BAQ = ∠BAP (l is bisector)

AB = AB (common)

Therefore, Δ APB ≅ Δ AQB by AAS congruence condition.

(ii) BP = BQ by Corresponding parts of congruent triangles are congruent, B is equidistant from the arms of ∠A

**In Fig. 7.21, AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE.**

Solution:

Given AC = AE , AB = AD and ∠BAD = ∠EAC

We have to prove, BC = DE

Let, ∠BAD = ∠EAC (Adding ∠DAC both sides)

∠BAD + ∠DAC = ∠EAC + ∠DAC

⇒∠BAC = ∠EAD

In, ΔABC and ΔADE,

AC = AE (given)

∠BAC = ∠EAD

AB = AD (Given)

Therefore, Δ ABC ≅ Δ ADE by SAS congruence condition

BC = DE by SAS congruence condition.

**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 (see Fig. 7.22). Show that**

**(i) Δ DAP ≅ Δ EBP**

**(ii) AD = BE**

Solution:

Given P is the midpoint of AB

∠BAD = ∠ABE and ∠EPA = ∠DPB

∠EPA = ∠DPB (adding ∠DPE both sides)

∠EPA + ∠DPE = ∠DPB + ∠DPE

⇒∠DPA = ∠EPB

In, Δ DAP ≅ Δ EBP,

∠DPA = EPB

AP = BP (P is the midpoint of AB)

∠BAD = ∠ABE (given)

Therefore, Δ DAP ≅ Δ EBP by ASA congruence condition

⇒AD = BE (Corresponding parts of congruent triangles are congruent)

**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 (see Fig. 7.23). Show that:**

**(i) Δ AMC ≅ Δ BMD**

**(ii) ∠ DBC is a right angle.**

**(iii) Δ DBC ≅ Δ ACB**

**(iv) CM = ^{1}/_{2} AB**

Solution:

(i) In ΔBMD and ΔAMC,

AM = BM (M is the midpoint)

∠CMA = ∠DMB (vertically opposite angles)

CM = DM (given)

⇒ ΔAMC ≅ ΔBMD by SAS congruence condition.

(ii) ∠ACM = ∠BDM (by Corresponding parts of congruent triangles are congruent)

Therefore, AC||BD as alternate interior angles are equal.

Now, ∠ACB + ∠DBC = 180˚

90˚ + ∠B = 180˚

∠DBC = 90˚

(iii) in ΔDBC and ΔACB

BC = CB (common)

∠ACB = ∠DBC (right angles)

DB = AC (byCorresponding parts of congruent triangles are congruent)

⸫ Δ DBC ≅ Δ ACB by SAS congruence condition.

(iv) DC = AB (ΔDBC ≅ ΔACB)

DM = CM = AM = BM (M is midpoint)

⇒ DM + CM = AM + BM

⇒ DM + CM = AB

⇒ CM = ^{1}/_{2} AB

Next exercise – Triangles – Exercise 7.2 – Class IX

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