# Math's: Division of a Line Segment: In a Ratio and Construction of Triangles

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## Division of a Line Segment in the Given Ratio Internally

To divide a line segment in a given ratio say , follow the steps given below-

**Step 1:**Draw a ray making an acute angle with .**Step 2:**Starting with , mark off points and on at equal distances from the point .**Step 3:**Join and .**Step 4:**Through (i.e. the second point), draw parallel to meeting at .

The point divides the line in the ratio

## Construction of Triangles

SSS Case (when three sides are given)- To construct a in which , and , follow the given steps

**Step 1:**Take any of the sides as base. Here, is taken as a base. Draw .**Step 2:**With as centre and radius , cut an arc.**Step 3:**With as centre and radius cut another arc intersecting the arc of Step 2 at .**Step 4:**Join and .

Then, is the required triangle**.**

SAS Case (when two sides and the included angle is given)- To construct a triangle in which , and

**Step 1:**Take one of the sides as base (e.g. take . Draw**Step 2:**At , construct an angle**Step 3:**With as centre and radius draw an arc cutting at .**Step 4:**Join

Then, is the required triangle.

ASA Case (when two angles and the included side is given)- To construct a triangle when two angles and one side are say , and , go through the following steps

**Step 1:**Draw .**Step 2:**At , construct**Step 3:**At , construct meeting at .

Then, ABC is the required triangle.

Right Triangle Case (when the hypotenuse and a side is given)- To construct a right triangle , right angled at , side and hypotenuse , follow the steps given below

**Step 1:**Draw**Step 2:**At , construct**Step 3:**With as centre and radius cut an arc intersecting at .**Step 4:**Join

is the required triangle.

When two sides and a median corresponding to one of these sides, are given- To construct a in which , and median , follow the given steps

**Step 1:**Draw**Step 2:**Draw the perpendicular bisector of meeting AB in D.**Step 3:**With as centre and radius cut an arc.**Step 4:**With as centre and radius cut another arc intersecting the arc of Step 3 at**Step 5:**Join AC and BC.

Then, is required triangle.