# 3. In Fig 7.33, $BD$ and $CE$ are altitudes of $\bigtriangleup ABC$ such that $BD= CE$.          (i) State the three pairs of equal parts in $\bigtriangleup CBD$ and $\bigtriangleup BCE$.           (ii) Is $\bigtriangleup CBD\cong \bigtriangleup BCE$ ? Why or why not?           (iii) Is $\angle DCB= \angle EBC$ ? Why or why not?

P Pankaj Sanodiya

i) Given,  in $\bigtriangleup CBD$ and $\bigtriangleup BCE$.

$BD= CE$

$\angle CEB=\angle BDC=90^o$

$\overline{ BC} = \overline{ CB}$

ii) So, By RHS Rule of congruency, we conclude:

$\bigtriangleup CBD\cong \bigtriangleup BCE$

iii) Since both the triangle are congruent, all parts of one triangle are equal to their corresponding part from another triangle.

So.

$\bigtriangleup CBD\cong \bigtriangleup BCE$.

Exams
Articles
Questions