Question

In triangle ABC, AD is perpendicular to BC and $A{D^2} = BD \times DC$ . Find $\angle BAC$
A. 90$^\circ $
B. 60$^\circ $
C. 30$^\circ $
D. 45$^\circ $

Answer 1

Hint: In this particular type of question we have to construct the diagram of triangle ABC with AD perpendicular to BC. When we have to use Pythagoras theorem in $\vartriangle ADB{\text{ and }}\vartriangle ADC$ and then further add both the equations to get the desired answer.

Complete step-by-step answer:

In triangle ABC, AD is perpendicular to BC and$A{D^2} = BD.DC$
In right triangles $\vartriangle ADB{\text{ and }}\vartriangle ADC$
Applying Pythagoras theorem,
\[
  A{B^2} = A{D^2} + B{D^2}{\text{ }}\left( i \right) \\
  A{C^2} = A{D^2} + D{C^2}{\text{ }}\left( {ii} \right) \\
\]
On adding both equations, we get
\[
  A{B^2} + A{C^2} = 2A{D^2} + B{D^2} + D{C^2} \\
   \Rightarrow A{B^2} + A{C^2} = 2BD.CD + B{D^2} + D{C^2}{\text{ }}\left( {{\text{Since }}A{D^2} = BD.CD} \right) \\
   \Rightarrow A{B^2} + A{C^2} = {\left( {BD + CD} \right)^2} = B{C^2} \\
\] (1)
Thus using the Converse of Pythagoras theorem in $\vartriangle ABC$
We get $\vartriangle ABC$is a right triangle and $\angle BAC = 90^\circ $

Note: Remember to recall the process of applying Pythagoras theorem while solving such types of questions. Also keep in mind that we have used the Converse of Pythagoras theorem at the last step of our solution. The converse of Pythagoras theorem states that if the square of the length of the longest side of a triangle is equal to the sum of the squares of the other two sides, then the triangle is a right triangle.