In the given figure $\angle BDC $ and $\angle BFC $ are right angles and $BF = CD$. By which congruence rule, can it be proved that $\Delta BDC $ is congruent to $\Delta CFB$

A) RHS B) SAS C) SSSS D) ASA
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Given $\Delta BDC$ and $\Delta BFC$ where $\angle BDC$ $=\angle BFC = 90°$ and $BF = CD$

To prove $\Delta BDC$ and $\Delta CFB$ are congruent by RHS rule

Solution:

In the triangles $\Delta BDC$ and $\Delta CFB$

$\angle BDC$ $=\angle BFC = 90^o$ (right angle)

$BC = BC$ (common Hypotenuse)

$BF = CD$ (side)

So by RHS Rule 

 $\Delta BDC$ $\cong$ $\Delta CFB$

So option A) RHS is CORRECT 

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Updated on: 10-Oct-2022

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