Punjab State Board PSEB 9th Class Maths Book Solutions Chapter 7 Triangles Ex 7.1 Textbook Exercise Questions and Answers.

## PSEB Solutions for Class 9 Maths Chapter 7 Triangles Ex 7.1

Question 1.

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

Answer:

In ∆ ABC and ∆ ABD,

AC = AD (Given)

∠ BAC = ∠ BAD (AB bisects ∠ A)

AB = AB (Common)

∴ ∆ ABC ≅ ∆ ABD (SAS rule)

∴ BC = BD (CPCT)

Thus, BC and BD are equal.

Question 2.

ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see the given figure). Prove that (i) ∆ ABD ≅ ∆ BAC, (ii) BD = AC and (iii) ∠ ABD = ∠ BAC

Answer:

In ∆ ABD and ∆ BAC,

AD = BC (Given)

∠ DAB = ∠ CBA (Given)

AB = BA (Common)

∴ ∆ ABD ≅ ∆ BAC (SAS rule)

∴ BD = AC (CPCT)

∴ ∠ ABD = ∠ BAC (CPCT)

Question 3.

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

Answer:

AD and BC are equal perpendiculars to line segment AB.

∴ AD = BC and ∠ OAD = ∠ OBC = 90°.

Now, in ∆ ADO and ∆ BCO,

AD = BC

∠ OAD = ∠ OBC

∠ AOD = ∠ BOC (Vertically opposite angles)

∴ ∆ ADO ≅ ∆ BCO (AAS rule)

∴ OA = OB (CPCT)

CD intersects AB at O and OA = OB.

Hence, CD bisects AB.

Question 4.

l and m are two parallel lines intersected by another pair of parallel lines p and q (see the given figure). Show that:

∆ ABC ≅ ∆ CDA.

Answer:

l || m and AC is their transversal.

∴ ∠ BCA = ∠ DAC (Alternate angles)

p l| q and AC is their transversal.

∴ ∠ BAC = ∠ DCA (Alternate angles)

Now, in ∆ ABC and ∆ CDA,

∠ BCA = ∠ DAC

∠ BAC = ∠ DCA

AC = CA (Common)

∴ ∆ ABC ≅ ∆ CDA (ASA rule)

Question 5.

Ray 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 the given figure). Show that:

(i) ∆ APB ≅ ∆ AQB

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

Answer:

l is the bisector of ∠ PAQ and B is any point on l.

∴ ∠ PAB = ∠ QAB

BP and BQ are perpendiculars from B to AP and AQ.

∴ ∠ BPA = ∠ BQA = 90°.

Now, in ∆ APB and ∆ AQB,

∠ PAB = ∠ QAB

∠ BPA = ∠ BQA

AB = AB (Common)

∴ ∆ APB ≅ ∆ AQB (AAS rule)

∴ BP = BQ (CPCT)

BP and BQ are perpendiculars from B to arms AP and AQ of ∠ A.

∴ BP is the distance of B from AP and BQ is the distance of B from AQ.

Thus, B is equidistant from the arms of ∠ A.

Question 6.

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

Answer:

∠ BAD = ∠ EAC

∴ ∠ BAD + ∠ DAC = ∠ EAC + ∠ DAC

∴ ∠ BAC = ∠ DAE (Adjacent angles)

Now, in ∆ BAC and ∆ DAE,

AC = AE (Given)

AB = AD (Given)

∠ BAC = ∠ DAE

∴ ∆ BAC ≅ ∆ DAE (SAS rule)

∴ BC = DE (CPCT)

Question 7.

AB is a line segment and P is its midpoint. D and E are points on the same side of AB such that ∠ BAD = ∠ ABE and ∠ EPA = ∠ DPB (see the given figure). Show that:

(i) ∆ DAP ≅ ∆ EBP

(ii) AD = BE

Answer:

∠ BAD = ∠ ABE

∴ ∠ PAD = ∠PBE (∵ P lies on AB.)

∠ EPA = ∠ DPB

∴ ∠ EPA + ∠ EPD = ∠ DPB + ∠ EPD

∴ ∠ APD = ∠ BPE (Adjacent angles)

P is the midpoint of AB.

∴ AP = BP

Now, in ∆ DAP and ∆ EBP

∠ PAD = ∠ PBE

∠ APD = ∠ BPE

AP = BP

∴ ∆ DAP ≅ ∆ EBP (ASA rule)

∴ AD = BE (CPCT)

Question 8.

In right triangle ABC, right angled at C, M is the midpoint 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 the given figure). Show that:

(i) ∆ AMC ≅ ∆ BMD

(ii) ∠ DBC is a right angle

(iii) ∆ DBC ≅ ∆ ACB

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

Answer:

In ∆ AMC and ∆ BMD,

AM = BM (∵ M is the midpoint of AB.)

CM = DM (Given)

∠ AMC = ∠ BMD (Vertically opposite angles)

∴ By SAS rule, ∆ AMC ≅ ∆ BMD [Result (i)]

∴ ∠ MCA = ∠ MDB (CPCT)

∠ MCA and ∠ MDB are alternate angles formed by transversal CD of lines AC and BD and they are equal.

∴ AC || BD

Now, ∠ DBC and ∠ ACB are interior angles on the same side of transversal BC of AC || BD.

∴ ∠ DBC + ∠ ACB = 180°

∴ ∠ DBC + 90° = 180° (Given : ∠ C = 90°)

∴ ∠ DBC = 90°

Thus, ∠ DBC is a right angle. [Result (ii)]

Now, ∆ AMC ≅ ∆ BMD

∴ AC = BD

In ∆ DBC and ∆ ACB,

BD = CA

∠ DBC = ∠ ACB (Right angles)

BC = CB (Common)

∴ ∆ DBC ≅ ∆ ACB [Result (iii)]

∴ DC = AB (CPCT)

DM = CM and M lies on line’ segment CD.

∴ DC = 2 CM

∴ AB = 2CM

∴ \(\frac{1}{2}\)AB = CM

∴ CM = \(\frac{1}{2}\)AB