Rough sketches of altitudes from A to BC for the given triangle are as follows:

The six elements i.e. the three sides and the three angles of \(\triangle ABC\) are as follows:
sides \(\bar{AB},\bar{BC},\bar{CA}\)
Angles \(\angle ABC,\angle BAC,\angle BCA\)

In the given figure, altitude can be drawn as below:

In right angled \(\triangle ABC\), altitude from A to BC will same as AB.

In the given figure, altitude can be drawn as below:

AL = Altitude from A to BC

In the adjacent figure, we have \(\triangle PQR\) We know that, a median connects a vertex of a triangle to the mid-point of the opposite side. On joining Q and mid-point of PR, i.e. E.We get the required median QE.

(i) \(\angle B=125^{0}\) which is greater than 90°, so it is an obtuse angled triangle.
(ii) All angles are smaller than 90°, so it is an acute angled triangle.
(iii) \(\angle B=90^{0}\), so it is a right angled rriangle.

(a) In the given figure,\(\bar { AC } =6cm,\bar { BC } =6cm\bar { ,AB } =4cm,\) \(\quad \because \bar { AC } =\bar { BC } \)
Hence, \(\triangle ABC\) is an isosceles triangle.
(b) In the given figure, \(\bar { AC } =4cm,\bar { AB } =4cm,\bar { BC } =4cm\)  \(\because \bar { AC } =\bar { AB } =\bar { BC } \)
Hence, \(\triangle ABC\) is an equilateral triangle.

According to the question,
Measure of \(\angle B=B\)
Measure of \(\angle A=B-40^0\)
Measure of \(\angle C=B-40^0+50^0\)
We know that, the sum of all three angles in a triangle is equal to 180°.
SO,(B - 40°) + B + (B - 40° + 50°) =180°
\(\Rightarrow\)3B - 30° =180°
\(\Rightarrow\)3B = 210°

\(\therefore B=\frac{210^0}{3}=70^0\)
So, measure of \(\angle A=70^0-40^0=30^0\)

According to the question,
Measure of \(\angle Y=Y\)
Measure of \(\angle X=Y+30^0\)
Measure of \(\angle Z=90^0\)
We know that, the sum of all three angles in a triangle is equal to 180°.
Y + (Y + 30°) + 90° = 180°
\(\Rightarrow\) 2Y + 120° =180°
\(\Rightarrow\) 2Y = 180° - 120°
\(\Rightarrow\) 2Y = 60°
\(\therefore y=\frac{60^0}{2}=30^0\)
Hence, X = 30° + 30° = 60°

In the given figure, QP II RT, where PR is a transversal line.
So, \(\angle x\) and \(\angle TRP\) are alternate angle.
\(\therefore\angle x=70^0\)
We know that, the sum of all angles in triangle is equal to 180°.
\(\therefore\angle x+30^0+\angle y=180^0\)
\(\Rightarrow\) 70° + 30° + Y =180°
\(\Rightarrow\) Y = 180° - 100°
\(\Rightarrow\) Y = 80°

Since, \(\angle Y\) and 45⁰ form a linear pair.
So, \(\angle y+45°=180°\)
\(\Rightarrow\angle y=180°-45°\)
\(\Rightarrow\angle y=135°\)
The sum of all angles in a triangle is equal to 180°.
So, \(45^0+60^0+\angle x=180°\)
\(\Rightarrow 105^0+\angle x=180°\)
\(\Rightarrow\angle x=180°-105°\)
=75°

Draw a line segment Be. By paper folding locate the perpendicular bisector of BC.The folded crease meets BC at D, its mid-point.
Take any point A on this perpendicular bisector. Join AB and Ae. Thus, the triangle obtained is an isoscelesMBC in which AB = AC.
Since, D is the mid-point of BC, so, AD is its median. Also, AD is perpendicular bisector of Be. So, AD is the altitude of \(\triangle ABC\).
Thus, it is verified that the median and altitude of an isoscelestriangle are same.

(a) In the following figure, we have \(\triangle ABC\).We know that, a median connects a vertex of a triangle to the mid-point of the opposite sides. On joining B and mid-point of AC i.e. E we get the required median BE.

(b) We know that, the perpendicular line segment from a vertex of a triangle to its opposite side is called an altitude of the triangle. In the following figure, we have \(\triangle PQR\) in which PQ and PR are the altitude drawn,from vertex Q to PR and from vertex R to PQ, respectively.

(c) In the adjoining figure, we have an obtuse angled triangle XYZ in which YL is an altitude drawn from Y to produced XZ such altitude lies in the exterior of the triangle.

A triangle has three vertices and three sides opposite to each of the vertices. So, for each vertex, there is a median of the triangle. Hence, a triangle has three medians.

In the above figure D, E and F are the mid-points of line segments BC, AC and AB, respectively. Therefore, line segments AD, BE and CF are three medians of \(\triangle ABC\).

(i) x + y = 1200                                  ...(1)
The exterior angle of a triangle is equal to the sum of its two interior opposite angles
x + y + y = 1800
Base angles opposite to the equal sides of an isosceles triangle are equal and the sum of the measures of the three angles of a triangle is 180⁰
\(\Rightarrow\)x + 2y = 1800                                  ... (2)
Subtracting equation (1) from equation (2),
y = 60°
Put y = 60° in equation (1),
x + 60° = 120°
\(\Rightarrow\)x = 120° - 60°
\(\Rightarrow\)x = 60°