On the basis of question we draw a figure of triangle ABC in which X and Y are the mid-points of sides AB and AC ,

The line joining the mid–points of two sides of a triangle is parallel to the third side and half of the base.

Correct Option: B

On the basis of question we draw a figure of triangle ABC in which X and Y are the mid-points of sides AB and AC ,

The line joining the mid–points of two sides of a triangle is parallel to the third side and half of the base.

According to question,
In a triangle ABC,
Given , ∠B = 5∠C; and ∠A = 3∠C
∴ ∠A + ∠B + ∠C = 180°
⇒ 3∠C + 5∠C + ∠C = 180°
⇒ 9∠C = 180°

Correct Option: C

According to question,
In a triangle ABC,
Given , ∠B = 5∠C; and ∠A = 3∠C
∴ ∠A + ∠B + ∠C = 180°
⇒ 3∠C + 5∠C + ∠C = 180°
⇒ 9∠C = 180°

As we know that the sum of the two sides of a triangle is greater than the third side.
For option, 11 cm., 3 cm., and 12 cm,
11 + 3 = 14 > 12
3 + 12 = 15 > 11
11 + 12 = 23 > 3

Correct Option: B

As we know that the sum of the two sides of a triangle is greater than the third side.
For option, 11 cm., 3 cm., and 12 cm,
11 + 3 = 14 > 12
3 + 12 = 15 > 11
11 + 12 = 23 > 3
Hence , option B is correct answer .

According to question , we draw a figure ∆ ABC

Given , ∠ABC = 35°
∠ACB = 65°
∴ ∠BAC + ∠ABC + ∠ACB = 180°
∠BAC = 180° – 35° – 65°
∠BAC = 180° – 100° = 80°

Correct Option: D

According to question , we draw a figure ∆ ABC

Given , ∠ABC = 35°
∠ACB = 65°
∴ ∠BAC + ∠ABC + ∠ACB = 180°
∠BAC = 180° – 35° – 65°
∠BAC = 180° – 100° = 80°
∠BAD = ∠DAC = 40°
∴ ∠ADB = 180° – 35° – 40° = 105°

If the orthocentre of a triangle lies on the side, it lies on the vertex.

Correct Option: A

If the orthocentre of a triangle lies on the side, it lies on the vertex. Hence , option A is correct answer .