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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?

Answers (1)

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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 $\bigtriangleup CBD\cong \bigtriangleup BCE$ .

Posted by

Pankaj Sanodiya

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