Question
On a common hypotenuse $AB,$ two right triangles $ACB$ and $ADB$ are situated on opposite sides. Prove that $\angle\text{BAC}=\angle\text{BDC}.$
✓
Answer
AB is the common hypotenuse of $\triangle\text{ACB}$ and $\triangle\text{ADB}.$
$\Rightarrow\ \angle\text{ACB}=90^\circ$ and $\angle\text{BDC}=90^\circ$
$\Rightarrow\ \angle\text{ACB}+\angle\text{BDC}=180^\circ$
$\Rightarrow $ The opposite angles of quadrilateral $ACBD$ are supplementary.
Thus, $ACBD$ is a cyclic quadrilateral.
This means that a circle passes through the points $A, C, B$ and $D.$
$\Rightarrow\ \angle\text{BAC}=\angle\text{BDC} [$angles in the same segment$]$
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