Question

There is a triangle $△ABC$ two perpendiculars which are equal in length are drawn to sides $AB$ and $AC$ from the vertices $C$ and $B$ respectively. Now, consider the statements below.

Which of the above statement is correct?

• A. $1$ and $2$ only
• B. $2$ and $3$ only
• C. $1$ and $3$ only
• D. $1$, $2$ and $3$

Answer 1

The correct option is D $1$, $2$ and $3$

In the figure,
In $△ABE$ and $△ACD$
$\angle A=\angle A$ (common in both)
$\angle AEB=\angle ADC\left({90}^o\right)$
$EB=DC$ (altitudes are equal)
Hence,$△ABE\cong △ACD$ (AAS)
So, $AB=AC$ (C.P.C.T)
hence, $△ABC$ isosceles triangle.

In $△BCD$ and $△CBE$
$BC=CB$ (common in both)
$\angle BDC=\angle CEB\left({90}^o\right)$
$CD=BE$ (equal altitudes)
So, $△BCD\cong △CBE$ (RHS)

So, all the statements are correct and option $\left(d\right)$ correct.