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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 similar triangles 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 corresponding sides' ratio is 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 corresponding sides' ratio is equal to the ratio of the corresponding perimeters' ratio.

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 equals 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, \(\frac{ar(\triangle ABD)}{ar(\triangle ADC)} = \frac{BD}{DC}\)