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Question

In the figure given below triangles $$ABC$$ is right-angled at $$B .\  ABPQ $$ and $$ACRS$$ are squares. Prove that :
$$ \Delta ACQ$$ and $$ \Delta ASB $$ are congruent

1841174_10bc848836c24ab98502c89455ec8fa6.png


Solution

Given ; A $$ \Delta ABC$$ is right-angled at $$B .$$
$$ABPQ$$ and $$PQRS$$ are square 
We need to prove that
$$ \Delta ACQ \cong \Delta ASB $$
$$ \angle QAB = 90^{0}$$
$$ \angle SAC = 90^{o}$$
$$\angle QAB = \angle SAC $$
Adding $$\angle BAC$$ to both sides of (3) , we have
$$\angle QAB  + \angle BAC = \angle SAC +\angle BAC $$
$$\Rightarrow  \angle QAC = \angle SAB $$  ...(1)

Now, In $$\Delta ACQ$$ and $$\Delta ASB$$,
$$QA=QB$$           ...[Sides of a square ABPQ]
$$\angle QAC=\angle SAB$$           ...[From (1)]
$$AC=AS  $$                 ....[Sides of a squre ACRS]
So, By angle-angle-side criterion of congruence,
$$\Delta ACQ \cong \Delta ASB$$

Mathematics

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