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

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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