### Theory:

Let us learn a few results on the similarity of triangles.

1.

A perpendicular line drawn from the vertex of a right angled triangle divides the triangle into two triangles similar to each other and the original triangle.

Here, $$\triangle ADB \sim \triangle ADC$$, $$\triangle BAC \sim \triangle BDA$$, $$\triangle BAC \sim \triangle ADC$$.

2.

If two triangles are similar, then the ratio of the corresponding sides are equal to the ratio of their corresponding altitudes.

That is, $$\triangle ABC \sim \triangle PQR$$, then $$\frac{AB}{PQ} = \frac{BC}{QR} = \frac{CA}{RP} = \frac{AD}{PS} = \frac{BE}{QT} = \frac{CF}{RU}$$

3.

If two triangles are similar, then the ratio of the corresponding sides are equal to the ratio of the corresponding perimeters.

That is, $$\triangle ABC \sim \triangle XYZ$$, then $$\frac{AB}{XY} = \frac{BC}{YZ} = \frac{CA}{ZX} = \frac{AB + BC + CA}{XY + YZ + ZX}$$

4.

The ratio of the area of two similar triangles are equal to the ratio of the squares of their corresponding sides.

That is, $$\frac{ar(\triangle ABC)}{ar(\triangle XYZ)} = \frac{AB^2}{XY^2} = \frac{BC^2}{YZ^2} = \frac{CA^2}{ZX^2}$$

5.

If two triangles have a common vertex and their bases are on the same straight line, the ratio between their areas is equal to the ratio between the length of their bases.

That is, $$\frac{ar(\triangle ABD)}{ar(\triangle ADC)} = \frac{BD}{DC}$$