Construction 11.4 :
To construct a triangle, given its base, a base angle and sum of other two sides.
Given the base BC, a base angle ∠B and the sum AB + AC,
we need to construct Δ ABC
Steps of Construction :
- Draw base BC and at point B make an angle XBC equal to the given angle.
- Cut a line segment BD equal to AB + AC from ray BX.
- Join DC. Draw perpendicular bisector PQ of line DC.
- Let PQ intersect BD a point A. Join AC.
Then, Δ ABC is the required triangle.
We need to prove that AB + AC = BD .
Given Base BC and ∠B
In Δ ACD,
PQ is the perpendicular bisector of CD
i.e. AR is the perpendicular bisector of CD
⇒ CR = DR
& ∠ ARC = ∠ ARD = 90°
Now, in Δ ADR and Δ ACR
AR = AR
∠ ARC = ∠ ARD
CR = DR
∴ Δ ADR ≅ Δ ACR
⇒ AC = AD
BD = AB + AD
BD = AB + AC
Thus, our construction is justified