Construct a ∆ABC in which AB = 5.8 cm, ∠B = 60° and BC + CA = 8.4 cm. Justify your construction.

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Solution

Steps of construction:
1. Draw a line segment AB = 5.8 cm.
2. Construct $∠$ ABX = 60 $°$ .
3. Set off BP = 8.4 cm.
4. Join PA.
5. Draw the right bisector of PA, meeting BP at C.
6. Join AC.
Thus, $△$ ABC is the required triangle.
Justification:
In $△$ APC,
∠CAP = ∠CPA (By construction)
⇒ CP = AC (Sides opposite to equal angles are equal)
Now
BC = PB − PC = PB − AC
⇒ BC + AC = PB

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