Construction 11.5 :
To construct a triangle, given its base, a base angle and the difference of the other two sides.
Let Δ ABC with ∠ B and base BC be given
Now, we are also given difference of other two sides
So, There can be two cases
AB > AC and difference AB – AC is given
AC > AB and difference AC – AB is given
Let’s do both these cases separately
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
2. Now, let’s draw ∠ B
Construct angle B from point B.
Let the ray be BX
3. Open the compass to length AB – AC.
From point B as center, cut an arc on ray BX.
Let the arc intersect BX at D
4. Join CD
Note: Since AB > AC, (AB – AC) is positive So, BD will be above line BC
Now, we will draw perpendicular bisector of CD
6. Mark point A where perpendicular bisector intersects BD
Join AC
∴ Δ ABC is the required triangle
Check Construction 11.2, Class 9 on how to draw perpendicular bisector
Justification – Case 1
We need to prove that AB – AC = BD.
Let perpendicular bisector intersect CD at point R
Thus,
AR is the perpendicular bisector of CD
∴ CR = DR
& ∠ ARC = ∠ ARD = 90°
Now,
In Δ ADR and Δ ACR
AR = AR
∠ ARD = ∠ ARC
DR = CR
∴ Δ ADR ≅ Δ ACR
⇒ AC = AD
Now,
BD = AB – AD
BD = AB – AC
Thus, our construction is justified
(Common)
(From (2))
(From (1))
(SAS Congruency)
(CPCT)
(From (3))
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
2. Now, let’s draw ∠ B
Construct angle B from point B.
Let the ray be BX
Open the compass to length AC – AB.
From point B as center, cut an arc on ray BX (opposite side of BC).
Let the arc intersect BX at D
4. Join CD
Note: Since AC > AB,
(AB – AC) is negative So, BD will be below line BC
Now, we will draw perpendicular bisector of CD
6. Mark point A where perpendicular bisector intersects BD
Join AC
∴ Δ ABC is the required triangle
Check Construction 11.2, Class 9 on how to draw perpendicular bisector
Justification - Case 2
We need to prove that AC – AB = BD.
Let perpendicular bisector intersect CD at point R
Thus,
AR is the perpendicular bisector of CD
∴ CR = DR
& ∠ ARC = ∠ ARD = 90°
Now,
In Δ ADR and Δ ACR
AR = AR
∠ ARD = ∠ ARC
DR = CR
∴ Δ ADR ≅ Δ ACR
⇒ AC = AD
Now,
BD = AD – AB
BD = AC – AB
Thus, our construction is justified
(Common)
(From (2))
(From (1))
(SAS Congruency)
(CPCT)

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.