Can someone help me prove these two triangles are similar?

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Answer 1 · 2018-02-12T03:29:57

As proved below

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Triangle ABC - A (2,1), B (6,3) C (8,1)

Triangle ADE - A (2,1), D (4,2), E (5,1)

Using distance formula,

$- - \to A B = \sqrt{{(6 - 2)}^{2} + {(3 - 1)}^{2}} = \sqrt{20}$ $vec(AB) = sqrt((6-2)^2 + (3-1)^2) = sqrt20$

$- - \to A C = \sqrt{{(8 - 2)}^{2} + {(1 - 1)}^{2}} = 6$ $vec(AC) = sqrt((8-2)^2 + (1-1)^2) = 6$

$- - \to A D = \sqrt{{(4 - 2)}^{2} + {(2 - 1)}^{2}} = \sqrt{5}$ $vec(AD) = sqrt((4-2)^2 + (2-1)^2) = sqrt5$

$- - \to A E = \sqrt{{(5 - 2)}^{2} + {(1 - 1)}^{2}} = 3$ $vec(AE) = sqrt((5-2)^2 + (1-1)^2) = 3$

$(AB) / (AD) = sqrt20 / sqrt5 = sqrt(4*5)/sqrt5 = (2 cancelsqrt5)/sqrt5 = 2$

$(AC) / (AE) = 6/3 = 2$

$(AB) / (AD) = (AC) / (AE) = 2$ and $hatA$ is common in both the triangles.

Hence $Delta ABC, Delta ADE$ are similar.

This can be confirmed by showing $(BC) / (DE)$ also $= 2$