**
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

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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.

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Justification
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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

Now,

BD = AB + AD

**
BD = AB + AC
**

Thus, our construction is justified