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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)

Davneet Singh has done his B.Tech from Indian Institute of Technology, Kanpur. He has been teaching from the past 13 years. He provides courses for Maths, Science, Social Science, Physics, Chemistry, Computer Science at Teachoo.

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