Triangles-9th-Math: 9 Math

Solution of NCERT Exercise 7.1: 9th math

Question (1) In quadrilateral ACBD, AC = AD and AB bisects ∠ A (see figure). Show that Δ ABC ≅ Δ ABD. What can you say about BC and BD?

Solution

Given, Quadrilateral ACBD,

and AC = AD

And AB bisects ∠ A

To prove

Δ ABC ≅ Δ ABD

And BC and BD =?

Proof

In Δ ABC and Δ ABD

AC = AD (Given)

∠ CAB = ∠ BAD

[Because according to question AB bisects ∠ A]

AB = AB [common in both of the triangle]

Therefore, by SAS (side-angle-side) congruence criterion

Δ ABC ≅ Δ ABD Proved

And now by CPCT (Corresponding parts of congruent triangles)

BC = BD

Thus, BC and BD are equal.

Question (2) ABCD is a quadrilateral in which AD = BC and ∠ DAB = ∠ CBA (see figure). Prove that

(i) Δ ABD ≅ &Delta BAC

(ii) BD = AC

(iii) ∠ ABD = ∠ BAC

Solution

Given, ABCD = a quadrilateral

AD = BC and

∠ DAB = ∠ CBA

To prove

(i) Δ ABD ≅ Δ BAC

Proof

In Δ ABD and Δ BAC

AD = BC (Given in question)

∠ DAB = ∠ CBA [As given in question]

AB = AB [Common side in both of the triangles]

Therefore, by SAS (side-angle-side) congruence criterion

Δ ABD ≅ Δ BAC Proved

(ii) BD = AC

As proved above, Δ ABD ≅ Δ BAC

Therefore, By CPCT (Corresponding parts of congruence triangles)

BD = AC Proved

(iii) ∠ ABD = ∠ BAC

As proved above, Δ ABD ≅ Δ BAC

Therefore, By CPCT (Corresponding parts of congruence triangles)

∠ ABD = ∠ BAC Proved

Question (3) AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.

Solution

Given, AD and BC are equal

And AD and BC are perpendiculars to AB

To prove

CD bisects AB i.e. OA = OB

Proof

In Δ BOC and Δ AOD

∠ OAD = ∠ OBC = 90⁰

BC = AD [As given in question]

∠ AOD = ∠ BOC [Because these are vertically opposite angles]

Therefore, by ASA (Angle-side-angle) congruence criterion

Δ BOC ≅ Δ AOD

Therefore, By CPCT (Corresponding parts of congruence triangles)

OA = OB Proved

Question (4) l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that Δ ABC ≅ Δ CDA.

Solution

Given, l and m are two parallel lines and intersected by another pair of parallel lines p and q

To prove

Δ ABC ≅ Δ CDA

Proof

In Δ ABC and Δ CDA

AC = AC [common side in both of the triangles]

∠ DAC = ∠ ACB and ∠ DCA = ∠ CAB

[Because, p and q is p are parallel and AC is the transversal, so these are alternate interior angles and hence are equal.]

Therefore, by AAS (Angle-Angle-Side) congruence criterion

Δ ABC ≅ Δ CDA Hence, proved

Question (5) Line l is the bisector of an angle ∠ A and B is any pint on l. BP and BQ are perpendiculars from B to the arms of ∠ A (see figure). Show that:

(i) Δ APB ≅ Δ AQB

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

Solution

Given, l bisects ∠ A,

and B is a point on l.

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

Therefore, To show

(i) Δ APB ≅ Δ AQB

Proof

In Δ ABP and Δ AQB

∠ BQA = ∠ BPA = 90⁰

AB = AB

(Common in both of the triangle)

∠ QAB = ∠ BAP

(Because as given in question, l bisects ∠ A)

Therefore, according to ASA (Angle-Side-Angle) congruence criterion

Δ ABP cong; Δ AQB Hence, proved

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

Proof

As proved above that, Δ ABP cong; Δ AQB

Thus, corresponding parts of congruence triangles are equal

Therefore, By CPCT (Corresponding parts of congruence triangles)

BP = BQ

That means B is equidistant from the arms of ∠ A Proved

Question (6) In figure AC = AE, AB = AD and ∠ BAD = ∠ EAC. Show that BC = DE

Solution

Given, AC = AE

AB = AD and

∠ BAD = ∠ EAC

To show

BC = DE

Proof

In Δ ABC and Δ ADE

AC = AE (As given in question)

AB = AD (As given in question)

∠ BAD = ∠ EAC (As given in question)

⇒ ∠ BAD + ∠ DAC = ∠ EAC + ∠ DAC

Now, according to Euclid's Geometry, we know that if equals are added to equals, the whole are equal

Therefore, ∠ BAC = ∠ EAD

Therefore, by SAS (Side-Angle-Side) congruence criterion

Δ AB ≅ Δ ADE

Therefore, By CPCT (Corresponding parts of congruence triangles)

BC = DE Hence proved

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 figure). Show that

(i) Δ DAP ≅ Δ EBP

(ii) AD = BE

Solution

Given, P is the mid-point of AB

∴ AP = BP

∠ BAD = ∠ ABE and

∠ EPA = ∠ DPB

To show,

(i) Δ DAP ≅ Δ EBP

(ii) AD = BE

Proof (i)

In Δ DAP and Δ EBP

AP = BP (Because P is the mid-point of AB as given in question)

∠ BAD = ∠ ABE (As given in the question)

∠ EPA = ∠ DPB

⇒ ∠ EPA + ∠ EPD = ∠ DPB + ∠ EPD

⇒ ∠ APD = ∠ EPB

[Because according to Euclid's Geometry, if equals are added to equals, the whole are equal]

Therefore, According to AAS (Angle-Angle-Side) congruence criterion

Δ DAP ≅ Δ EBP Hence, proved

Proof (ii)

Δ DAP ≅ Δ EBP (As proved above)

Now, by CPCT (Corresponding parts of congruence triangles)

AD = BE Proved

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) Δ ABC ≅ Δ BMD

(ii) ∠ DBC is a right angle

(iii) Δ DBC ≅ Δ ACB

(iv) CM = 1/2 AB

Solution

Given, Δ ABC is a right angle in which ∠ C=90⁰

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

DM = CM

To show

(i) Δ ABC ≅ Δ BMD

(ii) ∠ DBC is a right angle

(iii) Δ DBC ≅ Δ ACB

(iv) CM = 1/2 AB

Proof to show (i) Δ ABC ≅ Δ BMD

In Δ AMC and Δ BMD

DM = CM (As given in question)

BM = AM (As given in question, M is the mid-point of AB)

∠ BMD = ∠ AMC

(Vertically opposite angles. And we know that vertically opposite angles are equal)

Therefore, by SAS (Side-Angle-Side) congruence criterion

Δ AMC ≅ Δ BMD Proved

Proof to show (ii) ∠ DBC is a right angle

As proved above Δ AMC ≅ Δ BMD

Therefore, ∠ BDC = ∠ MCA

(∠ BDC and ∠ MCA are alternate interior angles and hence are equal)

Since alternate interior angles are equal,

Therefore, DB ‖ AC

Now, ∠ DBC + ∠ ACD = 180⁰

[Because sum of co-interior angles is equal to 180⁰. And ∠ DBC and ∠ ACD are co-interior angles ]

⇒ ∠ DBC + 90⁰ = 180⁰

∴ ∠ DBC = 180⁰ – 90⁰

⇒ ∠ DBC = 90⁰ Proved

Proof to show (iii) Δ DBC ≅ Δ ACB

In Δ DBC and Δ ACB

AC = BD

(Because, as proved above Δ AMC ≅ Δ BMD. Thus by CPCT (Corresponding parts of congruence triangles) AC = BD)

∠ MBD = ∠ MAC

(Because AC ‖ BD and ∠ MBD = ∠ MAC are alternate interior angles)

BC = BC (Common in both of the triangles)

Therefore, by SAS (Side-Angle-Side) congruence criterion

Δ DBC ≅ Δ ACB Hence, Proved

Proof to show (iv) CM = 1/2 AB

As proved above, Δ DBC ≅ Δ ACB

Now, by CPCT (Corresponding parts of congruence triangles)

AB = CD

Now, since M is the mid-point of AB

Therefore, 1/2 AB = 1/2 CD

⇒ 1/2 AB = CM

⇒ CM = 1/2 AB Hence, proved

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