Question

Using SAS (Side-angle-side postulate), find a point on a given line that is equally distant from two given points. Find the midpoint using SAS.

Answer 1

Let there are two points P , Q and a line l is given as shown.

We have to find a point on line l which is at equal distance from both point P and Q.

Step 1. Join the both point as shown

Step 2. Draw the perpendicular bisected of the line PQ.

Steps  to draw perpendicular bisector

1.) Take more then half distance of line PQ in compass.

2.) Make the arc on both upper and lower side of the line PQ from both the point P and Q such that they cut each other

3.) Let the cutting point of arc are A and B.

4.) Join the point A and B

5.) The line AB is perpendicular to line PQ.

Step 3. Let the cutting point of line l and AB is R.

And cutting point of line AB and PQ is M.

Hence

The point R is the required point

Proof.

Join the point R with P and Q.

Then two triangles are obtained triangle PRM and triangle RQM

In ∆ PRM and ∆ RQM

RP= RQ ( since AB is perpendicular bisector of PQ Angle P = Angle Q ( AB perpendicular to PQ )

PM= MQ ( AB is perpendicular to PQ )

So ∆ PRM Ξ ∆ RQM (SAS)

Hence R is the required point.

Similarly

Any point on line AB is at equal distance from point P and Q.