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 $\angle\text{BDC}=90^\circ$
$\Rightarrow\ \angle\text{ACB}+\angle\text{BDC}=180^\circ$
⇒ 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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