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
In right triangle ACB and ADB, we have
$\angle\text{ACB}=90^\circ$ and $\angle\text{ADB}=90^\circ$
$\therefore\angle\text{ACB}+\angle\text{ADB}=90^\circ+90^\circ=180^\circ$
If the sum of any pair of opposite angle of quadrilateral is 180°, then the quadrilateral is cyclic. So, ADBC is a cyclic quadrilateral.
Join CD. Angle $\angle\text{BAC}$ and $\angle\text{BDC}$ are made by $\widehat{\text{BC}}$ in the same segment BDAC.
Hence, $\angle\text{BAC}=\angle\text{BDC}.$
[$\because$ Angles in the same segment of a circle are equal]
$\angle\text{ACB}=90^\circ$ and $\angle\text{ADB}=90^\circ$
$\therefore\angle\text{ACB}+\angle\text{ADB}=90^\circ+90^\circ=180^\circ$
If the sum of any pair of opposite angle of quadrilateral is 180°, then the quadrilateral is cyclic. So, ADBC is a cyclic quadrilateral.
Join CD. Angle $\angle\text{BAC}$ and $\angle\text{BDC}$ are made by $\widehat{\text{BC}}$ in the same segment BDAC.
Hence, $\angle\text{BAC}=\angle\text{BDC}.$
[$\because$ Angles in the same segment of a circle are equal]
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