Jharkhand Board JAC Class 9 Maths Solutions Chapter 7 Triangles Ex 7.1 Textbook Exercise Questions and Answers.

## JAC Board Class 9th Maths Solutions Chapter 7 Triangles Ex 7.1

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Question 1.

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

Answer:

Given: AC = AD and AB bisects ∠A

To prove: ΔABC ≅ ΔABD

Proof: In AABC and AABD,

AB = AB (Common)

AC = AD (Given)

∠CAB = ∠DAB (AB is bisector)

Therefore, ΔABC ≅ ΔABD by SAS congruence criterion.

BC = BD (By CPCT).

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Question 2.

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

(i) ΔABD ≅ ΔBAC

(ii) BD = AC

(iii) ∠ABD = ∠BAC.

Answer:

Given: AD = BC and ∠DAB = ∠CBA

Proof:

(i) In ΔABD and ΔBAC,

AB = BA (Common)

∠DAB = ∠CBA (Given)

AD = BC (Given)

Therefore, ΔABD ≅ ΔBAC by SAS

congruence criterion.

(ii) Since, ΔABD ≅ ΔBAC Therefore, BD = AC (by CPCT)

(iii) Since, ΔABD ≅ ΔBAC Therefore, ∠ABD = ∠BAC (by CPCT)

Question 3.

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

Answer:

Given: AD and BC are perpendiculars to AB.

To prove: CD bisects AB i.e. OA = OB

Proof: In ΔAOD and ΔBOC,

∠A = ∠B = 90° (Perpendicular)

∠AOD = ∠BOC (Vertically opposite angles)

AD = BC (Given)

Therefore, ΔAOD ≅ ΔBOC by AAS congruence criterion.

Now, AO = OB (CPCT) i.e. CD bisects AB.

Question 4.

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

Answer:

Given: l || m and p || q

To prove: ΔABC ≅ ΔCDA

Proof: In ΔABC and ΔCDA,

As l || m

∴ ∠BCA = ∠DAC

(Alternate interior angles) AC = CA (Common)

As p 11q

∴ ∠BAC = ∠DCA (Alternate interior angles)

Therefore, ΔABC ≅ ΔCDA by ASA congruence criterion.

Question 5.

Line l is the bisector of an angle ∠A and B is any point on P. BP and BQ are perpendiculars from B to the arms of ∠A (See Fig.) Show that

(i) ΔAPB ≅ ΔAQB

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

Answer:

Given: l is the bisector of an angle ∠A.

BP and BQ are perpendiculars.

Proof:

(i) In AAPB and AAQB,

∠P = ∠Q (Right angles)

∠BAP = ∠BAQ (l is bisector)

AB = AB (Common)

Therefore, ΔAPB ≅ ΔAQB by AAS congruence criterion.

(ii) BP = BQ by CPCT.

(∵ ΔAPB ≅ ΔAQB)

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

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Question 6.

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

Answer:

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

To prove: BC = DE

Proof: ∠BAD = ∠EAC

∠BAD + ∠DAC = ∠EAC + ∠DAC (Adding ∠DAC both sides)

⇒ ∠BAC = ∠EAD

In AABC and AADE,

AC = AE (Given)

∠BAC = ∠EAD AB = AD (Given)

Therefore, ΔABC ≅ ΔADE by SAS congruence criterion.

BC = DE (by CPCT).

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). Show that

(i) ΔDAP ≅ ΔEBP

(ii) AD = BE

Answer:

Given: P is mid-point of AB.

∠BAD = ∠ABE and ∠EPA = ∠DPB

Proof:

(i) ∠EPA = ∠DPB

∠EPA + ∠DPE = ∠DPB + ∠DPE (Adding ∠DPE both sides)

⇒ ∠DPA = ∠EPB

In ΔDAP and ΔEBP,

∠DPA = ∠EPB

AP = BP (P is mid-point of AB)

∠BAD = ∠ABE (Given)

Therefore, ΔDAP ≅ ΔEBP by ASA congruence criterion.

(ii) 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 Fig). Show that:

(i) ΔAMC ≅ ΔBMD

(ii) ∠DRC is a right angle.

(iii) ΔDBC ≅ ΔACB

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

Answer:

Given: ∠C = 90°, M is the mid-point of AB and DM = CM

Proof:

(i) In ΔAMC and ΔBMD,

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

∠CMA = ∠DMB (Vertically opposite angles)

CM = DM (Given)

Therefore, ΔAMC ≅ ΔBMD by SAS congruence criterion.

(ii) ∠ACM = ∠BDM (by CPCT)

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

Now, ∠ACB + ∠DBC = 180° (co-interior angles)

⇒ 90° + ∠DBC = 180°

⇒ ∠DBC – 90°

(iii) In ΔDBC and ΔACB,

BC = CB (Common)

∠ACB = ∠DBC (Right angles)

DB = AC (by CPCT)

Therefore, ΔDBC ≅ ΔACB by SAS congruence criterion.

(iv) DC = AB (CPCT)

DM + CM = AB

⇒ CM + CM = AB (∵ DM = CM)

⇒ CM = AB

⇒ CM = \(\frac{1}{2}\) AB