 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$^\circ$

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.

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