## Triangles

**Chapter 7**

**Exercise 7.1**

**EX 7.1 QUESTION 1.**

**In quadrilateral ACBD, AC = AD and AB bisect A (see Fig. 7.16). Show that ΔABC ΔABD. What can you say about BC and BD?**

**Solution:**

In ∆ABC and ∆ABD,

AC = AD [Given]

∠ CAB = ∠ DAB [AB bisects ∠A]

and AB = AB [Common]

∴ ∆ ABC ≅ ∆ABD [SAS congruence rule]

∴ BC = BD [By CPCT]

**EX 7.1 QUESTION 2.**

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

**(i) ΔABD ΔBAC**

**(ii) BD = AC**

**(iii) ABD = BAC.**

**Solution:**

(i) In ∆ ABC and ∆ BAC,

AD = BC [Given]

∠DAB = ∠CBA [Given]

AB = AB [Common]

∴ ∆ ABD ≅ ∆BAC [By SAS congruence]

(ii) Since ∆ABD ≅ ∆BAC

⇒ BD = AC [By C.P.C.T.]

(iii) Since ∆ABD ≅ ∆BAC

⇒ ∠ABD = ∠BAC [By C.P.C.T.]

**EX 7.1 QUESTION 3.**

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

**Solution:**

In ∆OCB and ∆ODA, we have

∠CBO = ∠DAO [Each 90°]

BC = AD [Given]

∠BOC = ∠AOD [Vertically opposite angles]

∴ ∆OCB ≅ ∆ODA [By AAS congruency]

⇒ BO = AO [By C.P.C.T.]

CD bisects AB.

**EX 7.1 QUESTION 4.**

*l *and *m* are two parallel lines intersected by another pair of parallel lines p and q (see Fig. 7.19). Show that ΔABC ΔCDA.

**Solution:**

∠BAC = ∠ACD [Alternate angles]

CA = AC [Common]

∠BCA = ∠DAC [Alternate angles]

∴ ∆ABC ≅ ∆CDA [By ASA congruency rule]

**EX 7.1 QUESTION 5.**

**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) ΔAPB ΔAQB**

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

**Solution:**

In ∆APB and ∆AQB

∠ABP = ∠AQB [Proved above]

AB = BA [Common]

∠PAB = ∠QAB [Given]

∴ ∆APB ≅ ∆AQB [By ASA congruency]

(ii) BP = BQ [By C.P.C.T.]

**EX 7.1 QUESTION 6.**

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

**Solution:**

∠BAD = ∠EAC [Given]

Adding ∠DAC on both sides, we have

∠BAD + ∠DAC = ∠EAC + ∠DAC

⇒ ∠BAC = ∠EAD

Now, in ∆BAC and ∆DAE.

∠BAC = ∠DAE [Proved above]

AB = AD [Given]

AC = AE [Given]

∴ ∆ABC ≅ ∆ADE [By SAS congruency]

⇒ BC = DE [By C.P.C.T.]

**EX 7.1 QUESTION 7.**

**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) ΔDAP ΔEBP**

**(ii) AD = BE**

**Solutions:**

(i)∠EPA = ∠DPB [Given]

Adding ∠EPD on both sides, we get

∠EPA + ∠EPD = ∠DPB + ∠EPD

⇒ ∠APD = ∠BPE

(i)In ∆DAP and ∆EBP,

∠A = ∠B [Given]

AP = BP [P is the mid-point of AB.]

∠APD = ∠BPE [Proved above]

∴ ∆DAP ≅ ∆EBP [By ASA congruency]

(ii) AD = BE [By C.P.C.T.]

**EX 7.1 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 Fig. 7.23). Show that:**

**(i) ΔAMC ΔBMD**

**(ii) DBC is a right angle.**

**(iii) ΔDBC ΔACB**

**(iv) CM = ½ AB**

**Solution:**

M is the midpoint of AB.

∴ BM = AM

(i) In ∆AMC and ∆BMD,

CM = DM [Given]

∠AMC = ∠BMD [Vertically opposite angles]

AM = BM [Proved above]

∴ ∆AMC ≅ ∆BMD [By SAS congruency]

(ii) Since ∆AMC ≅ ∆BMD

⇒ ∠CAM = ∠DBM [By C.P.C.T.]

alternate interior angles (∠CAM = ∠DBM) are equal,

∴ AC || DB

Now, BC is a transversal which intersects parallel lines AC and DB,

∴ ∠ACB + ∠DBC = 180° [Co-interior angles]

∠ACB = 90° [∆ABC is right angled at C]

∴ 90° + ∠DBC = 180°

⇒ ∠DBC = 90°

(iii) ∆AMC ≅ ∆BMD [Proved above]

∴ AC = BD [By C.P.C.T.]

In ∆DBC and ∆ACB,

BD = CA [Proved above]

∠DBC = ∠ACB [Each 90°]

BC = CB [Common]

∴ ∆DBC ≅ ∆ACB [By SAS congruency]

(iv) As ∆DBC ≅ ∆ACB

DC = AB [By C.P.C.T.]

But DM = CM [Given]

∴ CM = ** ½**DC = ** ½**AB

⇒ CM = ** ½**AB.