Answer :

Steps of construction:

i. Now in order to make a triangle, draw a line segment AB = 8 cm.

ii. Draw two arcs intersecting at 4 cm distance from points A and B; on either side of AB.

Join these arcs to get perpendicular bisector CD of AB. (Since, altitude is the perpendicular bisector of base of isosceles triangle).

iii. Join points A and B to C in order to get the triangle ABC.

iv. Now, draw a ray AX which is at an acute angle from point A.

As

And 3 is greater between 3 and 2,

So

Plot 3 points on AX such that:

AA_{1} = A_{1}A_{2} = A_{2}A_{3}.

v. As 2 is smaller between 2 and 3. Join A_{2} to point B.

Draw a line from A_{3 }which is parallel to A_{2}B meeting the extension of AB at B’.

vi. Draw B’C’ || BC. Then, draw A’C’ || AC.

Triangle AB’C’ is the required triangle.**Justification:**We need to prove,

By construction

As C

^{'}B

^{'}|| CB

They will maker equal angles with line AB.

∠ACB = ∠AC

^{'}B

^{'}..... (corresponding angles)

In ΔACB and ΔAC

^{'}B

^{'}

∠A = ∠A (common)

∠ACB = ∠AC

^{'}B

^{'}(corresponding angles)

So ΔACB ∼ ΔAC

^{'}B

^{'}

As corresponding sides of similar triangles are in ratio,

Hence,

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