\(\displaystyle\angle{B}={88}^{{\circ}}\)

\(\displaystyle\angle{D}={49}^{{\circ}}\)

\(\displaystyle\angle{C}={180}^{{\circ}}-{\left({88}^{{\circ}}+{49}^{{\circ}}\right)}\)

\(\displaystyle={43}^{{\circ}}\)

In the triangle IJH,

\(\displaystyle\angle{I}={43}^{{\circ}}\)

\(\displaystyle\angle{J}={49}^{{\circ}}\)

\(\displaystyle\angle{H}={180}^{{\circ}}-{\left({43}^{{\circ}}+{49}^{{\circ}}\right)}\)

\(\displaystyle={88}^{{\circ}}\)

Then, \(\displaystyle\angle{B}=\angle{H},\angle{D}=\angle{J},\angle{C}=\angle{I}\)

So, two triangles BCD and IJH will be similar if \(\displaystyle{\frac{{{B}{D}}}{{{H}{J}}}}={\frac{{{B}{C}}}{{{H}{I}}}}={\frac{{{C}{D}}}{{{I}{J}}}}\)

Now,

\(\displaystyle{\frac{{{B}{D}}}{{{H}{J}}}}={\frac{{{6}}}{{{4}}}}\)

\(\displaystyle={\frac{{{3}}}{{{2}}}}\)

\(\displaystyle{\frac{{{B}{C}}}{{{H}{I}}}}={\frac{{{9}}}{{{6}}}}\)

\(\displaystyle={\frac{{{3}}}{{{2}}}}\)

\(\displaystyle{\frac{{{C}{D}}}{{{I}{J}}}}={\frac{{{12}}}{{{8}}}}\)

\(\displaystyle={\frac{{{3}}}{{{2}}}}\)

Since \(\displaystyle{\frac{{{B}{D}}}{{{H}{J}}}}={\frac{{{B}{C}}}{{{H}{I}}}}={\frac{{{C}{D}}}{{{I}{J}}}}\), triangle BCD and HIJ are similar.