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
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