# Ex.6.5 Q3 Triangles Solution - NCERT Maths Class 10

## Question

In Figure, \({ABD}\) is a triangle right angled at \(A\) and \({AC⊥BD}\). Show that

(i) \(A{{B}^{2}}=\text{ }BC.\text{ }BD\)

(ii) \(A{{C}^{2}}=\text{ }BC.\text{ }DC\)

(iii) \(A{{D}^{2}}=\text{ }BD.\text{ }CD\)

**Diagram**

## Text Solution

**Reasoning:**

As we know if a perpendicular is drawn from the vertex of the right angle of a right triangle to the hypotenuse then triangles on both sides of the perpendicular are similar to the whole triangle and to each other.

**Steps:**

i). In \(\Delta BAD\;\text{and}\;\Delta BCA\)

\[\begin{array}{l}

\Rightarrow \,\,\Delta BAD \sim \Delta BCA\\

\Rightarrow \frac{{AB}}{{BC}} = \frac{{BD}}{{AB}}{\rm{ }}\qquad\left( {{\text{Corresponding sides of similar triangle}}} \right)\\

\Rightarrow A{B^2} = BC \cdot BD

\end{array}\]

ii). In \(\Delta BCA\,\,\text{and}\,\Delta ACD\)

\[\begin{array}{l}

\Rightarrow \,\Delta BCA \sim \Delta ACD\\

\, \Rightarrow \,\frac{{AC}}{{CD}} = \frac{{BC}}{{AC}}{\rm{ }}\qquad\;\left( {{\text{Corresponding sides of similar triangle}}} \right)\\

\Rightarrow \,A{C^2} = BC \cdot DC

\end{array}\]

iii). In \(\Delta BAD\) and \(\Delta ACD\)

\[\begin{array}{l}

\Rightarrow \Delta BAD \sim \Delta ACD\\

\Rightarrow \frac{{AD}}{{CD}} = \frac{{BD}}{{AD}}{\rm{ }}\qquad\;\left( {{\text{Corresponding sides of similar triangle}}} \right)\\

\,\,\,\,\,A{D^2} = BD \cdot CD

\end{array}\]