Steps:
- Draw a line segment AB of any given length.
- Construct ∠BAX and ∠ABY at A and B such that ∠A = ∠B
- \(\overrightarrow {AX}\) and \(\overrightarrow {BY}\) will intersect at C.
- ΔABC is the required triangle.
Justification:
Drop a perpendicular CM to AB from C.
Now in ΔAMC and ΔBMC
∠AMC = ∠BMC [Right angle]
∠A = ∠B [Construction]
CM = CM (Common)
∴ ΔAMC ≅ ΔBMC
⇒ AC = BC [CPCT]