Prove triangle similiarity by given expression I am working on the following problem, but I can't seem to figure it out.

The length of the sides in the triangle $T_1$ are $a_1$, $b_1$ and $c_1$. The length of the sides in the triangle $T_2$ are $a_2$, $b_2$ and $c_2$. Moreover: $$\sqrt{a_1 a_2} + \sqrt{b_1 b_2} + \sqrt{c_1 c_2} = \sqrt{(a_1 + b_1 +c_1)(a_2 + b_2 +c_2)}.$$ Show that the triangles are similar.

Any help would be appreciated

It's easy. Assume that, $a_i$'s, $b_i$'s and $c_i$'s are corresponding sides.
Now squaring both sides, and cancelling common terms both side, you get the identity something like this,$$\sum\limits_{\text{cyclic}}(\sqrt{a_1b_2}-\sqrt{b_1a_2})^2=0.$$
Which means, each square term is individually $0$. So, $$\sqrt{a_1b_2}-\sqrt{b_1a_2}=0\\\implies\frac{a_1}{a_2}=\frac{b_1}{b_2}.$$
Do this same approach for other two summands.
Which ultimately yields, $$\frac {a_1}{a_2}=\frac {b_1}{b_2}=\frac {c_1}{c_2},$$ proving the similarity of those triangles.
And it is the criteria for equality in Cauchy-Schwarz inequality.
