**
Construction
**
**
11.5 :
**

To construct a triangle, given its base, a base angle and the difference of the other two sides.

There can be two cases

- AB > AC and difference AB – AC is given
- AC > AB and difference AC – AB is given

__
Case 1
__
: AB > AC

Given the base BC, a base angle ∠B and the difference 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 BX a point A. Join AC.

Then, Δ ABC is the required triangle.

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Justification
__

We need to prove that
**
AB – AC = BD
**
.

Given Base BC and ∠B

In Δ ACD,

PQ is the perpendicular bisector of CD

∴ AD = AC

Now,

BD = AB – AD

**
BD = AB
**
**
–
**
**
AC
**

Thus, our construction is justified

__
Case 2
__
: AC > AB

Given the base BC, a base angle ∠B and the difference AC – AB,

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 BX a point A. Join AC.

Then, Δ ABC is the required triangle.

__
Justification
__

We need to prove that
**
AC
**
**
–
**
**
AB
**
**
= BD
**
.

Given Base BC and ∠B

In Δ ACD,

PQ is the perpendicular bisector of CD

∴ AD = AC

Now,

BD = AB – AD

**
BD = AB – AC
**

Thus, our construction is justified