If the triangle is not required to be congruent, it is actually bisecting an edge 16, and the line from the bisecting point to the vertex of the corresponding angle. These lines divide the triangle into 16 small triangles with equal areas. Because the base and height of these small triangles are equal.

If these small triangles are required to be congruent. Then all right triangles will do.

First, divide the big triangle into four congruent middle triangles.

A is a right angle. AD is the median line on BC, which divides △ABC into two isosceles triangles ADB and ADC with equal size (because the base is equal and the height is * * *), and AD=BD=DC (the median line on the hypotenuse of a right triangle is equal to half of the hypotenuse. )

Then draw vertical lines DE and DF from d to AB and AC respectively, because the height on the bottom of isosceles triangle is also the center line. So the triangle ADB and ADC are divided into two congruent triangles respectively. These four triangles BDE, DEA, DAF and DFC are congruent, and their areas are equal. And they are all right triangles.

Then according to the practice just now, the four right-angled triangles BDE, DEA, DAF and DFC can all be divided into four congruent small right-angled triangles. In this way, the total * * * is divided into 4×4= 16 congruent right triangles.