MHB Prove Similar Triangles: $\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1}$

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Two triangles with sides a, b, c and a1, b1, c1 are proven to be similar if and only if the equation √(aa1) + √(bb1) + √(cc1) equals √((a+b+c)(a1+b1+c1)). This condition establishes a direct relationship between the sides of the triangles, indicating similarity. The proof involves manipulating the equation to demonstrate the equivalence of the two triangle configurations. The discussion emphasizes the importance of this relationship in geometry. Understanding this concept is crucial for further studies in triangle properties and similarity.
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Prove that two triangles with sides $a,\,b,\,c$ and $a_1,\,b_1,\,c_1$ are similar if and only if $\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1}=\sqrt{(a+b+c)(a_1+b_1+c_1)}$.
 
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anemone said:
Prove that two triangles with sides $a,\,b,\,c$ and $a_1,\,b_1,\,c_1$ are similar if and only if $\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1}=\sqrt{(a+b+c)(a_1+b_1+c_1)}$.
we have
$\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1}=\sqrt{(a+b+c)(a_1+b_1+c_1)}$.

$\equiv (\sqrt{aa_1}+\sqrt{bb_1}+\sqrt{cc_1})^2=(a+b+c)(a_1+b_1+c_1)$

$\equiv aa_1+bb_1+cc_1+2\sqrt{aa_1bb_1} + 2\sqrt{bb_1cc_1} + 2\sqrt{cc_1aa_1} = aa_1+ab_1 + ac_1 + ba_1 + bb_1 + bc_1 + ca_1 + cb_1 + cc_1$

$\equiv 2\sqrt{aa_1bb_1} + 2\sqrt{bb_1cc_1} + 2\sqrt{cc_1aa_1} = ab_1 + ac_1 + ba_1 + bc_1 + ca_1 + cb_1$

$\equiv ab_1 + ac_1 + ba_1 + bc_1 + ca_1 + cb_1-2(\sqrt{aa_1bb_1} + 2\sqrt{bb_1cc_1} + 2\sqrt{cc_1aa_1}) = 0$

$\equiv (\sqrt{ab_1} - \sqrt{a_1b})^2 + (\sqrt{ac_1} - \sqrt{a_1c})^2 + (\sqrt{bc_1} - \sqrt{b_1c})^2 = 0$

The above is true iff $ab_1 = a_1b$, $ac_1 = a_1c$, $bc_1 = b_1c$

giving $\frac{a}{a_1} = \frac{b}{b_1} = \frac{c}{c_1}$ or the 2 triangles are similar
 
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