NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1 are part of NCERT Solutions for Class 9 Maths. Here we have given NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1.

## NCERT Solutions for Class 9 Maths Chapter 5 Triangles Ex 5.1

**Exercise 5.1**

**Question 1.**

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

**Solution:**

In ∆ABC and ∆ABD, we have

AC = AD (Given)

∠ CAB = ∠ DAB (∵ AB bisects ∠A)

and AB = AB (Common)

∴ ∆ ABC ≅ ∆ABD (By SAS congruence axiom)

∴ BC = BD (By CPCT)

**Question 2.**

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

**(i) ∆ABD ≅ ∆BAC**

**(ii) BD = AC**

**(iii) ∠ABD = ∠ BAC**

**Solution:**

In ∆ ABC and ∆ BAC, we have

AD = BC (Given)

∠DAB = ∠CBA (Given)

and AB = AB (Common)

∴ ∆ ABD ≅ ∆BAC (By SAS congruence axiom)

Hence, BD = AC (By CPCT)

and ∠ABD= ∠BAC (By CPCT)

**Question 3.**

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

**Solution:**

In ∆AOD and ∆BOC, we have

∠AOD = ∠BOC

∵ AB and CD intersects at O.

∴ Which are vertically opposite angle

∵ ∠DAO = ∠CBO = 90°

and AD = BC (Given)

∴ ∆AOD ≅ ∆BOC (By SAS congruence axiom)

⇒ O is the mid-point of AB.

Hence, CD bisects AB.

**Question 4.**

**l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ∆ABC = ∆CDA. **

**Solution:**

From figure, we have

∠ 1 = ∠ 2 (Vertically opposite angles).. .(i)

∠ 1 = ∠ 6 (Corresponding angles)…(ii)

∠ 6 = ∠ 4 (Corresponding angles) …(iii)

From Eqs. (i) (ii) and (iii), we have

∠ 1 = ∠ 4

and ∠ 2 = ∠ 4 …..(iv)

In ∆ABC and ∆CDA, we have

∠ 4 = ∠ 2 [From Eq. (iv]

∠5 = ∠ 3 (Alternate interior angles)

and AC = AC (Common)

∴ ∆ABC ≅ ∆ CDA (By AAS congruence axiom)

**Question 5.**

**Line l is the bisector of an ∠ A and ∠ B is any point on l. BP and BQ are perpendiculars from B to the arms of LA (see figure). Show that**

**(i) ∆APB ≅ ∆AQB**

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

**Solution:**

In ∆ APB and ∆AQB, we have

∠ APB = ∠ AQB = 90°

∠ PAB = ∠ QAB (∵ AB bisects ∠ PAQ)

and AB = AB (Common)

∴ ∆ APB ≅ ∆ AQB (By AAS congruence axiom)

⇒ BP = BQ (By CPCT)

⇒ B is equidistant from the arms of ∠ A.

**Question 6.**

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

**Solution:**

In ∆ABC and ∆ADE, we have

AB = AD (Given)

∠ BAD = ∠ EAC (Given)…(i)

On adding ∠ DAC on both sides in Eq. (i)

⇒ ∠ BAD + ∠ DAC = ∠ EAC + ∠ DAC

⇒ ∠ BAC = ∠ DAE

and AC = AE (Given)

∴ ∆ABC ≅ ∆ADE (By AAS congruence axiom)

⇒ BC = DE (ByCPCT)

**Question 7.**

**AS 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 figure). Show that**

**(i) ∆DAP ≅ ∆EBP**

**(ii) AD = BE **

**Solution:**

We have,

AP = BP [∵ P is the mid-point of AB (Given)]… (i)

∠ EPA = ∠ DPB (Given)…(ii)

∠ BAD = ∠ ABE (Given).. .(iii)

On adding ∠ EPD on both sides in Eq. (ii); we have

⇒ ∠ EPA + ∠ EPD = ∠DPB + ∠ EPD

⇒ ∠ DP A = ∠ EPB …..(iv)

Now, In ∆DAP and ∆EBP, we have

∠ DPA = ∠ EPB [ From Eq(4)]

∠ DAP = ∠ EBP (Given)

and AP = BP [From Eq. (i)]

∴ ∆ DAP ≅ ∆ EBP (By ASA congruence axiom)

Hence, AD = BE (By CPCT)

**Question 8.**

**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 figure). Show that**

**(i) ∆AMC ≅ ∆BMD**

**(ii) ∠DBC is a right angle**

**(iii) ∆DBC ≅ ∆ACB **

**(iv) CM = \(\frac { 1 }{ 2 }\) AB**

**Solution:**

Given ∆ACB in which ∠C = 90° and M is the mid- point of AB.

To prove (i) ∆ AMC = ∆ BMD

(ii) ∠DBC is a right angle

(iii) ∆DBC ≅ ∆ACB

(iv) CM = \(\frac { 1 }{ 2 }\) AB

Construction Produce CM to D, such that CM = MD. Join DB.

Proof In ∆ AMC and ∆ BMD, we have

AM = BM (M is the mid-point of AB)

CM = DM (Given)

and ∠AMC=∠BMD (Vertically opposite angles)

∴ ∆ AMC ≅ ∆ BMD (By SAS congruence axiom)

⇒ AC = DB (By CPCT) …(i)

and ∠1 = ∠2 (By CPCT)

Which are alternate angles

∴ BD || CA

Now, BD || CA and BC is transversal

∴ ∠ ACB + ∠CBD =180°

⇒ 90°+ ∠CBD = 180°

⇒∠CBD =90°

⇒ ∠ DBC = 90° [Which is part (ii)]

In ∆ DBC and ∆ ACB, we have

CB = BC (Common)

DB = AC [Using part (i)]

and ∠ CBD = ∠ BCA (Each 90°)

∴ ∆ DBC ≅ ∆ACB (By SSA congruence axiom) [Which is part (iii)]

⇒ DC=AB (by CPCT)

⇒ \(\frac { 1 }{ 2 }\) DC = \(\frac { 1 }{ 2 }\) AB

⇒ CM = \(\frac { 1 }{ 2 }\) AB (∵ CM = \(\frac { 1 }{ 2 }\) DC) [ Which is part(iv)]

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