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Last updated at May 29, 2023 by Teachoo
Example 2 Construct a triangle similar to a given triangle ABC with its sides equal its to 5/3 of the corresponding sides of the triangle ABC (scale factor 5/3). Here, we are given Ξ ABC, and scale factor 5/3 β΄ Scale Factor > 1 We need to construct triangle similar to Ξ ABC Letβs follow these steps Steps of construction Draw any ray BX making an acute angle with BC on the side opposite to the vertex A. Mark 5 (the greater of 5 and 3 in 5/3) points π΅_1,π΅_2,π΅_3,π΅_4,π΅_5on BX so that γπ΅π΅γ_1=π΅_1 π΅_2=π΅_2 π΅_3=π΅_3 π΅_4 =π΅_4 π΅_5 Join π΅_3C (3rd point as 3 is smaller in 5/3) and draw a line through π΅_5 parallel to π΅_3 πΆ, to intersect BC extended at Cβ². Draw a line through Cβ² parallel to the line CA to intersect BA extendedat Aβ². Thus, Ξ Aβ²BCβ² is the required triangle Justification Since scale factor is 5/3, we need to prove (π¨^β² π©)/π¨π©=(π¨^β² πͺ^β²)/π¨πͺ=(π©πͺ^β²)/π©πͺ = π/π. By construction, BC^β²/π΅πΆ=(π΅π΅_5)/(π΅π΅_3 )= 5/3. Also, AβCβ is parallel to AC So, they will make the same angle with line BC β΄ β AβCβB = β ACB Justification Since scale factor is 5/3, we need to prove (π¨^β² π©)/π¨π©=(π¨^β² πͺ^β²)/π¨πͺ=(π©πͺ^β²)/π©πͺ = π/π. By construction, BC^β²/π΅πΆ=(π΅π΅_5)/(π΅π΅_3 )= 5/3. Also, AβCβ is parallel to AC So, they will make the same angle with line BC β΄ β AβCβB = β ACB (Corresponding angles) Now, In Ξ AβBCβ and Ξ ABC β B = β B (Common) β AβCβB = β ACB (From (2)) Ξ AβBCβ βΌ Ξ ABC (AA Similarity) Since corresponding sides of similar triangles are in the same ratio (π΄^β² π΅)/π΄π΅=(π΄^β² πΆ^β²)/π΄πΆ=(π΅πΆ^β²)/π΅πΆ So, (π¨^β² π©)/π¨π©=(π¨^β² πͺ^β²)/π¨πͺ=(π©πͺ^β²)/π©πͺ = π/π. (From (1)) Thus, our construction is justified