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 angle equal to ∠ B from point X
Let the angle be ∠ LXY.
Make angle equal to ∠ C from point Y
Let the angle be ∠ MYX
4. Bisect ∠ LXY and ∠ MYX.
Let these bisectors intersect at a point A .
5. Make perpendicular bisector of AX
Let it intersect XY at point B
6. Make perpendicular bisector of AY
Let it intersect XY at point C
7. Join AB & AC
∴ Δ ABC is the required triangle
Justification
To justify, we have to prove
AB + BC + AC = XY
∠ LXY = ∠ B
∠ MYX = ∠ C
Now,,
Point B lies in perpendicular bisector of AX
∴ BX = BA
Similarly,
Point C lies in perpendicular bisector of AX
∴ CY = CA
Steps of Construction :
Draw a line segment, say XY equal to BC + CA + AB.
Make angles LXY equal to ∠B and MYX equal to ∠C.
Bisect ∠LXY and ∠MYX. Let these bisectors intersect at a point A .
For proof of BX = BA, see Construction 11.4
For proof of CY = CA, see Construction 11.4
Now, we know that
XY = XQ + BC + CY
XY = AB + BC + AC
So, XY = AB + BC + AC is proved,
now let’s prove ∠ LXY = ∠ B & ∠ MYX = ∠ C
(From (1) & (2))
In Δ AXB
Since BX = BA
∴ ∠ BAX = ∠ AXB
Now,
∠ ABC is the exterior angle of triangle AXB
∠ ABC = ∠ BAX + ∠ AXB
∠ ABC = ∠ AXB + ∠ AXB
(Angles opposite to equal sides are equal)
(Exterior angle is sum of interior opposite angles)
∠ ABC = 2 ∠ AXB
∠ ABC = ∠ LXY
∴ ∠ B = ∠ LXY
Similarly, we can prove
∠ A = ∠ MYX
Hence justified
As AX is bisector of ∠ LXY,
∴ ∠ AXB = (∠ 𝐿𝑋𝑌)/2
2∠ AXB = ∠ LXY

Made by

Davneet Singh

Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. He has been teaching from the past 12 years. He provides courses for Maths and Science at Teachoo.