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**Q. (1) In two right triangles one side an acute angle of one are equal to the corresponding side and angle of the other. Prove that the triangles are congruent.**

** Solution:**

Given that, in two right triangles one side and acute angle of one are equal to the corresponding side and angles of the other.

We have to prove that the triangles are congruent.

Let us consider two right triangles such that

âˆ B = âˆ C = 90^{âˆ˜} â€¦â€¦.(i)

AB = DE â€¦â€¦.(ii)

âˆ C = âˆ F From(iii)

Now observe the two triangles ABC and DEF

âˆ C = âˆ F [From (iii)]

âˆ B = âˆ E [From (i)]

and AB = DE [From (ii)]

So, by AAS congruence criterion, we have Î” ABC â‰… Î” DEF

Therefore, the two triangles are congruent

Hence proved

** **

**Q. (2) If the bisector of the exterior vertical angle of a triangle be parallel to the base. Show that the triangle is isosceles.**

**Solution:**

Given that the bisector of the exterior vertical angle of a triangle is parallel to the base and we have to prove that the triangle is isosceles. Let ABC be a triangle such that AD is the angular bisector of exterior vertical angle EAC and AD âˆ¥ BC

Let âˆ EAD = (i), âˆ DAC = (ii), âˆ ABC = (iii) and âˆ ACB = (iv)

** **

We have,

(i) = (ii) [AD is a bisector of âˆ EAC]

(i) = (iii) [Corresponding angles]

and (ii) = (iv) [alternative angle]

(iii) = (iv)

- AB = AC

Since, in Î” ABC, two sides AB and AC are equal we can say that Î” ABC is isosceles triangle.

**Q. (3) In an isosceles triangle, if the vertex angle is twice the sum of the base angles, calculate the angles of the triangle.**

**Solution:**

Let Î” ABC be isosceles such that AB = AC.

- âˆ B = âˆ C

Given that vertex angle A is twice the sum of the base angles B and C. i.e., âˆ A = 2(âˆ B + âˆ C)

âˆ A = 2(âˆ B + âˆ B)

âˆ A = 2(2 âˆ B)

âˆ A = 4(âˆ B)

Now, We know that sum of angles in a triangle = 180Â°

âˆ A + âˆ B + âˆ C = 180Â°

4 âˆ B + âˆ B + âˆ B = 180Â°

6 âˆ B = 180Â°

âˆ B = 30Â°

Since, âˆ B = âˆ C

- âˆ B = âˆ C = 30Â°

And âˆ A = 4 âˆ B

- âˆ A = 4 x 30Â° = 120Â°

Therefore, angles of the given triangle are 120Â°, 30Â° and 30Â°.

= 428 and LB = LC]

** **

**Q. (4) PQR is a triangle in which PQ = PR and is any point on the side PQ. Through S, a line is drawn parallel to QR and intersecting PR at T. Prove that PS = PT.**

**Solution :** Given that PQR is a triangle such that PQ = PR ant S is any point on the side PQ and ST âˆ¥ QR.

** **

To prove,

PS = PT

Since, PQ = PR

- PQR is an isosceles triangle.
- âˆ Q = âˆ R (or) âˆ PQR = âˆ PRQ

Now, âˆ PST = âˆ PQR and âˆ PTS = âˆ PRQ [Corresponding angles as ST parallel to QR]

Since, âˆ PQR = âˆ PRQ

- âˆ PST = âˆ PTS

Now, In Î” PST, âˆ PST = âˆ PTS

Î” PST is an isosceles triangle

Therefore, PS = PT

**Q. (5) In a Î” ABC, it is given that AB = AC and the bisectors of B and C intersect at O. If M is a point on BO produced, prove that âˆ MOC = âˆ ABC.**

**Solution:**

Given that in Î” ABC,

AB = AC and the bisector of âˆ B and âˆ C intersect at O. If M is a point on BO produced

** **

We have to prove âˆ MOC = âˆ ABC

Since,

AB = AC

- ABC is isosceles
- âˆ B = âˆ C (or)
- âˆ ABC = âˆ ACB

Now,

BO and CO are bisectors of âˆ ABC and âˆ ACB respectively

= > ABO = âˆ OBC = âˆ ACO = âˆ OCB = âˆ ABC = âˆ ACB â€¦â€¦(i)

We have, in Î” OBC

âˆ OBC + âˆ OCB + âˆ BOC = 180Â° â€¦â€¦(ii)

And also

âˆ BOC + âˆ COM = 180Â° â€¦â€¦(iii) [Straight angle]

Equating (ii) and (iii)

- âˆ OBC + âˆ OCB + âˆ BOC = âˆ BOC + âˆ MOC
- âˆ OBC + âˆ OCB = âˆ MOC [From (i)]
- 2 âˆ OBC = âˆ MOC [From (i)]
- 2 ( âˆ ABC) = âˆ MOC [From (i)]
- âˆ ABC = âˆ MOC

Therefore, âˆ MOC = âˆ ABC

**Q. (6) P is a point on the bisector of an angle ABC. If the line through P parallel to AB meets BC at Q, prove that triangle BPQ is isosceles.**

**Sol:**

Given that P is a point on the bisector of an angle ABC, and PQ âˆ¥ AB.

** **

We have to prove that Î” BPQ is isosceles.

Since,

BP is bisector of âˆ ABC = âˆ ABP = âˆ PBC â€¦â€¦(i)

Now,

PQ âˆ¥ AB

- âˆ BPQ = âˆ ABP â€¦â€¦(ii) [alternative angles]

From (i) and (ii), we get

âˆ BPQ = âˆ PBC (or) âˆ BPQ = âˆ PBQ

Now,

In BPQ,

âˆ BPQ = âˆ PBQ

Î” BPQ is an isosceles triangle.

Hence proved

**Q. (7) Prove that each angle of an equilateral triangle is 60Â°.**

**Sol:**

Given to prove each angle of an equilateral triangle is 60Â°.

Let us consider an equilateral triangle ABC.

Such that AB = BC = CA

Now, AB = BC

- âˆ A = âˆ C â€¦â€¦(i) [Opposite angles to equal sides are equal]

And BC = AC

- âˆ B = âˆ A â€¦â€¦(ii)

From (i) and (ii), we get

âˆ A = âˆ B = âˆ C â€¦â€¦(iii)

** **

We know that

Sum of angles in a triangle = 180

- âˆ A + âˆ B + âˆ C = 180
- âˆ A + âˆ A + âˆ A = 180
- 3 âˆ A = 180
- âˆ A = 60

âˆ A = âˆ B = âˆ C = 60

Hence, each angle of an equilateral triangle is 60Â°.

**Q. (8) Angles Î” A, B, C of a triangle ABC are equal to each other. Prove that ABC is equilateral.**

**Sol:**

Given that angles A, B, C of a triangle ABC equal to each other.

** **

We have to prove that Î” ABC is equilateral

We have, âˆ A = âˆ B = âˆ C

Now,

âˆ A = âˆ B

- BC = AC [opposite sides to equal angles are equal]

And âˆ B = âˆ C

- AC = AB

From the above we get

AB = BC = AC

- Î” ABC is equilateral.

**Q. (9) ABC is a triangle in which âˆ B = 2 âˆ C. D is a point on BC such that AD bisects âˆ BAC and AB = CD. Prove that [âˆ BAC = 72** **.**

**Solution:** Given that in ABC, âˆ B = 2 âˆ C and D is a point on BC such that AD bisectors âˆ BAC and

AB = CD.

** **

We have to prove that âˆ BAC = 72Â°

Now, draw the angular bisector of âˆ ABC, which meets AC in P.

Join PD

Let C = âˆ ACB = y

- âˆ B = âˆ ABC = 2 âˆ C = 2y and also
- Let âˆ BAD = âˆ DAC
- âˆ BAC = 2x [ AD is the bisector of âˆ BAC ]

Now, in Î” BPC,

- âˆ CBP = y [BP is the bisector of âˆ ABC ]
- âˆ PCB = y
- âˆ CBP = âˆ PCB = y [PC = BP]

Consider, Î” ABP and Î” DCP, we have

Î” ABP = Î” DCP = y

AB = DC [Given]

And PC = BP [From above]

So, by SAS congruence criterion, we have Î”ABPâ‰…Î”DCP

Now,

âˆ BAP = âˆ CDF and AP = DP [Corresponding parts of congruent triangles are equal]

- âˆ BAP = âˆ CDP = 2

Consider, Î” APD,

We have AP = DP

= âˆ ADP = âˆ DAP

But âˆ DAP = x

- âˆ ADP = âˆ DAP = x

Now

In Î” ABD.

âˆ ABD + âˆ BAD + âˆ ADB = 180

And also âˆ ADB + âˆ ADC = 180Â° [Straight angle]

From the above two equations, we get

âˆ ABD + âˆ BAD + âˆ ADB = âˆ ADB + âˆ ADC

- 2y + x = âˆ ADP + âˆ PDC
- 2y + x = x + 2x
- 2y = 2x
- y = x (or) x = y

We know,

Sum of angles in a triangle = 180Â°

So, In Î” ABC,

- âˆ A + âˆ B + âˆ C = 180Â°
- 2x + 2y + y = 180Â° [âˆ A = 2x, âˆ B = 2y, âˆ C = y]
- 2(y)+3y = 180Â° [x = y]
- 5y = 180Â°
- y = 36Â°

Now, âˆ A = âˆ BAC = 2 x 36Â° = 72Â°

**Q. (10) ABC is a right angled triangle in which âˆ A = 90Â° and AB = AC. Find âˆ B and âˆ C.**

**Solution:** Given that ABC is a right angled triangle such that âˆ A = 90Â° and AB = AC

Since, AB = AC

- Î” ABC is also isosceles.

Therefore, we can say that Î” ABC is right angled isosceles triangle.

- âˆ C = âˆ B and âˆ A = 90Â° â€¦â€¦(i)

Now, we have sum of angled in a triangle = 180Â°

- âˆ A + âˆ B + âˆ C = 180Â°
- 90Â° + âˆ B + âˆ B = 180Â° [From(i)]
- 2âˆ B = 180Â° â€“ 90Â°
- âˆ B = 45Â°

Therefore, âˆ B = âˆ C = 45Â°

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