Question

In$\Delta $ABC, AB = AC and BC = 6cm. D is a point on the side AC such that AD = 5 cm and CD = 4 cm. Show that $\Delta $BCD$\sim \Delta $ACB and hence find BD.

Answer 1

Hint: In triangle ABC, AB and AC are equal. The length of BC, AD and CD along with the angles $\angle ABC=\angle ACB $ are given. Now, using this given data we can prove both the triangles are similar and once this proof is verified then we can easily calculate the length of BD.

Complete step-by-step answer:

Given: The length of AB and AC are equal. The length of BC is 6cm. The length of AD is 5cm. The length of CD is 4cm

AB=AC
BC = 6cm.
AD = 5cm.
CD = 4cm.
To prove: $\Delta ACB\sim \Delta BCD$. Also find the length of BD.
Proof: $AB=AC$ (Given)
$\angle ABC=\angle ACB$ (Given)
The sum of AD and DC is equal to AC.
$\begin{align}
  & AD+DC=5cm+4cm \\
 & AC=9cm \\
\end{align}$
$\begin{align}
  & AB=9cm \\
 & BC=6cm \\
\end{align}$

Now, we have to prove that both the triangles are similar.
In $\Delta ACB\text{ and }\Delta \text{BCD}$
$\begin{align}
  & \angle ACB=\angle BCD...(1) \\
 & \dfrac{CB}{CD}=\dfrac{6}{4}=\dfrac{3}{2} \\
 & \dfrac{AC}{BC}=\dfrac{9}{6}=\dfrac{3}{2} \\
 & \dfrac{CB}{CD}=\dfrac{AC}{BC}...(2) \\
\end{align}$

From the equation (1) and equation (2) using the angle side relation,
$\Delta ACB\sim \Delta BCD$

Both the above triangles are similar.

Now, to find out BD we use the similarity ratio between corresponding sides of a similar triangle.

As, $\begin{align}
  & \dfrac{CB}{CD}=\dfrac{AB}{BD} \\
 & \therefore BD=\dfrac{CD}{CB}\times AB \\
 & BD=\dfrac{9\times 4}{6}=6cm \\
\end{align}$

Therefore, the length of BD is 6 cm.

Note: The key step in solving this problem is the knowledge of similar triangle properties and their proof. To prove a triangle is similar we required two justifying equations. After proving the triangle similar we can use the ratio of corresponding sides of the triangle to obtain the required length. Students must know the difference between congruence and similar triangles.