Prove that in similar triangles the ratio of two similar sides is equal to the ratio of the two corresponding medians.

| March 31, 2021 | education

Given: triangle ABC is similar to triangle A1B1C1, AM is the median, A1M1 is the median.
Prove: AB: A1B1 = AM: A1M1.
Evidence. The ABC triangle is similar to the A1B1C1 triangle by condition, therefore, similar sides are proportional and similar angles are equal. Angle B = angle B1, AB: A1B1 = BC: B1C1.
AM is the median by condition, therefore BM = CM = 0.5 BC.
A1M1 is the median by condition, therefore B1M1 = C1M1 = 0.5 B1C1.
ВМ: В1М1 = 0.5 ВС: (0.5 В1С1) = ВС: В1С1.
Consider triangles ABM and A1B1M1, they are similar in the second sign of similarity, which means that the similar sides are proportional: AB: A1B1 = AM: A1M1.
Conclusion: in similar triangles, the ratio of two similar sides is equal to the ratio of the two corresponding medians.



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