**Construction** **11.6 ****: **

To construct a triangle, given its perimeter and its two base angles.

Given the base angles, ∠B and ∠C and AB + AC + BC, we have to construct ΔABC.

** Steps of Construction** :

- Draw a line segment XY equal to AB + AC + BC.
- Make ∠ LXY equal to ∠B and ∠ MYX equal to ∠C.
- Bisect ∠ LXY and ∠ MYX. Let these bisectors intersect at a point A .
- Draw perpendicular bisectors of line AX and AY
- Let PQ intersect XY at B and RS intersect XY at C.
- Join AB and AC

Thus, Δ ABC is a the required triangle

__Justification__

To justify, we have to prove

- AB + BC + AC = XY
- ∠ LXY = ∠ B
- ∠ MYX = ∠ C

**In ****Δ**** AXQ**,

PQ is the perpendicular bisector of AX,

∴ BX = BA

Similarly, we can say

CY = CA

Now, we know that

XY = **XQ** + BC + **CY**

XY =** AB** + BC +** AC**

**In ****Δ**** AXQ**

Since AX = BA

∴ ∠ BAX = ∠ AXB

Now, ∠ ABC is the exterior angle of triangle AXQ

∠ ABC = ∠ BAX + ∠ AXB

∠ ABC = ∠ AXB + ∠ AXB

∠ ABC = 2 ∠ AXB

∠ ABC = ∠ LXY

Thus, ∠ B = ∠ LXY

Similarly, we can prove ∠ A = ∠ MYX

Hence justified