Question
In a rectangle $ABCD,$ prove that $\triangle\text{ACB}=\triangle\text{CAD}$.
✓
Answer
In rectangle $ABCD, AC$ is its diagonal.
Now in $\triangle\text{ACB}$ and $\triangle\text{CAD}$
$\text{AB}=\text{CD}$ (Opposite sides of a rectangle)
$\text{BC}=\text{AD}$
$\text{AC}=\text{AC}$ (Common)
$\triangle\text{ACB}=\triangle\text{CAD}$ (SSS condition)
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