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Example 1
Important
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Constructions

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Last updated at May 29, 2023 by Teachoo

Construction 11.5 : To construct a triangle, given its base, a base angle and the difference of the other two sides. Let Ξ ABC with β 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 β AC is given AC > AB and difference AC β AB is given Letβs do both these cases separately Case 1: AB > AC Given the base BC, a base angle β B and the difference AB β AC, we need to construct Ξ ABC Steps of Construction: Draw base BC 2. Now, letβs draw β B Construct angle B from point B. Let the ray be BX 3. Open the compass to length AB β 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 β 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 β΄ Ξ ABC is the required triangle Check Construction 11.2, Class 9 on how to draw perpendicular bisector Justification β Case 1 We need to prove that AB β AC = BD. Let perpendicular bisector intersect CD at point R Thus, AR is the perpendicular bisector of CD β΄ CR = DR & β ARC = β ARD = 90Β° Now, In Ξ ADR and Ξ ACR AR = AR β ARD = β ARC DR = CR β΄ Ξ ADR β Ξ ACR β AC = AD Now, BD = AB β AD BD = AB β 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 β B and the difference AC β AB, we need to construct Ξ ABC Steps of Construction: Draw base BC 2. Now, letβs draw β B Construct angle B from point B. Let the ray be BX Open the compass to length AC β 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 β 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 β΄ Ξ 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 β AB = BD. Let perpendicular bisector intersect CD at point R Thus, AR is the perpendicular bisector of CD β΄ CR = DR & β ARC = β ARD = 90Β° Now, In Ξ ADR and Ξ ACR AR = AR β ARD = β ARC DR = CR β΄ Ξ ADR β Ξ ACR β AC = AD Now, BD = AD β AB BD = AC β AB Thus, our construction is justified (Common) (From (2)) (From (1)) (SAS Congruency) (CPCT)