Triangles-9th-Math: 9 Math

Solution of NCERT Exercise 7.3: 9th math

Question (1) Δ ABC and Δ DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (See figure). If AD is extended to intersect BC at P, show that

(i) Δ ABD ≅ Δ ACD

(ii) Δ ABP ≅ Δ ACP

(iii) AP bisects ∠ A as well as ∠ D

(iv) AP is the perpendicular bisector of BC

Solution

Given, Δ ABC and Δ DBC are two isosceles triangles on the same base BC. A and D are vertices and are on same side of BC

To show

(i) Δ ABD ≅ Δ ACD

Proof

In Δ ABD and &Delta ACD

Both are isosceles triangles

Therefore, AB = AC

[As Δ ABC is isosceles triangle]

AD = AC (Common in both of the triangle)

BD = DC (As Δ BDC is an isosceles triangle)

Therefore, by SSS congruence criterion

Δ ABD ≅ Delta; ACD Hence, proved

(ii) Δ ABP ≅ Δ ACP

As proved above Δ ABD ≅ Delta; ACD

Therefore, by CPCT (Corresponding parts of congruence triangles)

∠ BAD = ∠ DAC -----------(i)

Now, In Δ ABP and Δ ACP

∠ BAP = ∠ PAC [As proved in equation (i)]

AP = AP [Common in both of the triangle]

∠ ABP = ∠ ACP

[Since, Δ ABC is an isosceles triangle and angles opposite to equal sides are equal]

Therefore, by ASA congruence criterion

Δ ABP ≅ Δ ACP

Hence, proved

(iii) AP bisects ∠ A as well as ∠ D

From equation (i)

∠ BAD = ∠ DAC

⇒ ∠ BAP = ∠ PAC

⇒ AP bisects angle A

And in Δ BPD and Δ PDC

BD = DC

(As Δ BDC is an isosceles triangle)

DP = DP (Common in both of the triangle)

BP = PC

[Because &Delat; ABP ≅ APC, By CPCT]

Therefore, by SSS congruence criterion

Δ BPD ≅ Δ PDC

And hence,

By CPCT (Corresponding parts of congruence triangles)

∠ BDP = ∠ PDC

Therefore, AP bisects ∠ A as well as ∠ D also

(iv) AP is the perpendicular bisector of BC

As Δ BPD ≅ Δ PDC (Proved above)

By CPCT (Corresponding parts of congruence triangles)

∠ BPD = ∠ CPD

And BP = CP

Now, since BC is straight line,

Thus, ∠ BPD + ∠ CPD = 180⁰

⇒ ∠ BPD + ∠ BPD = 180⁰

⇒ 2 ∠ BPD = 180⁰

⇒ ∠ BPD = 180⁰ / 2

Therefore, ∠ BPD = 90⁰ Hence, proved

Question (2) AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

(i) AD bisects BC

(ii) AD bisects ∠ A

Solution

Let, ABC is the given isosceles triangle

Given,

In this triangle ABC

AB = AC and AD is the altitude

Therefore, to prove

(i) AD bisects BC

In triangle ADB and triangle ADC

AB = AC (As given in question)

AD = AD (Common side in both of the triangle)

Therefore, by SAS congruence criterion

Δ ADB ≅ Δ ADC

And by CPCT (Corresponding parts of congruence triangles)

BD = DC

Therefore, AD bisects BC proved

(ii) AD bisects ∠ A

As proved above Δ ADB ≅ Δ ADC

By CPCT (Corresponding parts of congruence triangles)

&angl; BAD = &angl DAC

Therefore, AD bisects angle A Proved

Question (3) Two sides AB and BC and median AM of one triangle ABC are respectively equal to sides PQ and QR and median PN of Δ PQR (See figure). Show that

(i) Δ ABM ≅ Δ PQN

(ii) Δ ABC ≅ Δ PQR

Solution

Given, ABC and PQR are two triangles in which

Side AB = side PQ and side BC = side QR

and median AM of Δ AMC = median PN of Δ PQR

Therefore, to show

(i) Δ ABM ≅ Δ PQN

In Δ ABM and Δ PQN

AB = PQ (As given in question)

AM = PN (As given in question that medians of both of the triangles are equal)

BM = QN

[Since, BC = QR

⇒ 1/2 BC = 1/2 QR

⇒ BM = QN]

Therefore, by SSS congruence criterion

Δ ABM ≅ Δ PQN Hence, proved

Proof (ii) Δ ABC ≅ Δ PQR

Now, in Δ ABC and Δ PQR

AB = PQ (As given in question)

And since Δ ABM ≅ PQN, thus, By CPCT (Corresponding parts of congruence triangles)

∠ ABC = ∠ PQR

BC = QR (As given in question)

Thus, according to SAS congruence criterion

Δ ABC ≅ Delta; PQR Proved

Question (4) BE and CF are two equal altitudes of a triangle ABC. Using RHS congruence rule, prove that the triangle is isosceles.

Solution

Let ABC is the given triangle

In which given, BE = CF = altitudes

To prove: Δ ABC is an isosceles triangle Using RHS rule,

Proof

In Δ CBF and Δ BEC

∠ CFB = ∠ BEC = 90⁰

[Because as given in question, BE and CF are altitudes of triangle ABC]

BC = BC (Common side in both of the triangle)

BE = CF (As given in question, altitudes are equal)

Therefore, by RHS congruence criterion

Δ CBF ∠ Δ BEC

Therefore, by CPCT (Corresponding parts of congruence triangles)

∠ FBC = ∠ ECF

⇒ AB = AC

(Because, in an isosceles triangle angles opposite to equal sides are equal)

Therefore, Δ ABC is an isosceles triangle Proved

Question (5) ABC is an isosceles triangle with AB = AC. Draw AP ⊥ BC to show that ∠ B = ∠ C.

Solution

Given, Δ ABC is an isosceles triangle with AB = AC and AP ⊥ BC

To show ∠ B = ∠ C

Proof:

In Δ ABP and Δ ACP

AB = AC (As given in the question)

AP = AP (Common side in both of the triangle)

∠ APB = ∠ APC = 90⁰

(Because, as per question, AP ⊥ BC)

Therefore, by RHS congruence criterion

Δ APB ≅ ACP

Therefore, by CPCT (Corresponding parts of congruence triangles)

∠ B = ∠ C Hence, proved

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