2021/066) 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