MCQ
In the adjoining figure, $AC = BD$. If $\angle\text{CAB} = \angle\text{DBA},$ then $\angle\text{ACB}$ is equal to:
- ✓$\angle\text{BDA}$
- B$\angle\text{BAD}$
- C$\angle\text{ABC}$
- D$\angle\text{ABD}$
✓
Answer
Correct option: A.
$\angle\text{BDA}$
In Triangle $CAB$ and traingle $DBA,$
$AC = BD$ and $\angle\text{CAB} = \angle\text{DBA}$ (Given)
$AB$ (Common)
Therefore, Triangle $CAB$ and traingle $DBA$ are congruent by $SAS$ criteria
Therefore, $\angle\text{ACB} = \angle\text{BDA}$ (by $CPCT)$
$AC = BD$ and $\angle\text{CAB} = \angle\text{DBA}$ (Given)
$AB$ (Common)
Therefore, Triangle $CAB$ and traingle $DBA$ are congruent by $SAS$ criteria
Therefore, $\angle\text{ACB} = \angle\text{BDA}$ (by $CPCT)$
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