### Theory:

**:**

*Illustration*Consider a triangle \(ABC\) right angled at \(A\) with its hypotenuse \(BC\) at its base.

Draw a perpendicular to the triangle as follows:

Two smaller right triangles \(ABD\) and \(ACD\) are obtained.

Now, all the three triangles \(ABC\), \(ADB\) and \(ADC\) are similar.

Based on this similarity, the following theorem is obtained.

*Statement*:

If a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then:

**(i)**Triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

That is, \(\Delta ABC \sim \Delta ADB \sim \Delta ADC\).

**(ii)**\(AD^2 = BD \cdot DC\)

**(iii)**\(AC^2 = BC \cdot DC\) and \(AB^2 = BC \cdot BD\) where \(BC = BC + DC\)

*Proof for the theorem*:

**Proof**

*of**statement**(i):*

The perpendicular divides the triangle \(ABC\) into two right triangles \(ADB\) and \(ADC\).

Thus, \(\angle ABC\) \(=\) \(\angle ADB\) \(=\) \(\angle ADC\).

Consider \(\Delta ABC\) and \(\Delta ADB\).

The \(\angle B\) is common to both the triangles.

Also, given that \(\angle ABC\) \(=\) \(\angle ADB\) \(=\) \(90^{\circ}\).

Thus by AA similarity (If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar), it is concluded that the triangles \(ABC\) and \(ADB\) are similar.

That is, \(\Delta ABC\) \(\sim\) \(\Delta ADB\). \(…… (1)\).

Consider \(\Delta ABC\) and \(\Delta ADC\).

The \(\angle C\) is common to both the triangles.

Also, given that \(\angle ABC\) \(=\) \(\angle ADC\) \(=\) \(90^{\circ}\).

Thus by AA similarity (If two angles of one triangle are respectively equal to two angles of another triangle, then the two triangles are similar.), it is concluded that the triangles \(ABC\) and \(ADC\) are similar.

That is, \(\Delta ABC\) \(\sim\) \(\Delta ADC\). \(…… (2)\).

From equations \((1)\) and \((2)\), we have:

\(\Delta ADB\) \(\sim\) \(\Delta ADC\) \(…… (3)\).

From equations \((1)\), \((2)\) and \((3)\) it can be concluded that:

\(\Delta ABC\) \(\sim\) \(\Delta ADB\) \(\sim\) \(\Delta ADC\)

Therefore, the triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

**Proof of statement**(ii):

From the statement (i) we have \(\Delta ADB\) \(\sim\) \(\Delta ADC\).

Since the triangles are similar, the ratios of the corresponding sides of the triangles are also equal.

Thus we have \(\frac{BD}{AD} = \frac{AD}{DC}\).

This implies \(AD^2 = BD \cdot DC\).

**Proof of statement**(iii):

From the statement (i) we have \(\Delta ADC\) \(\sim\) \(\Delta ABC\).

Since the triangles are similar, the ratios of the corresponding sides of the triangles are also equal.

Thus we have \(\frac{AC}{BC} = \frac{DC}{AC}\).

This implies \(AC^2 = DC \cdot BC\).

Similarly, consider the similar triangles \(\Delta ADB\) \(\sim\) \(\Delta ABC\).

Since the triangles are similar, the ratios of the corresponding sides of the triangles are also equal.

Thus we have \(\frac{AB}{BC} = \frac{DB}{AB}\).

This implies \(AC^2 = DC \cdot BC\).

Example:

From the figure, find the altitude \(h\).

Solution:

By the statement (ii) of the theorem, the altitude is computed as follows:

\(h^2 = BD \times DC\)

\(h^2 = 4 \times 9\)

\(h^2 =\) \(36\)

\(\Rightarrow h = \sqrt{36}\)

\(= 6\)

Therefore, the measure of the altitude is \(6\) \(cm\).