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 $180^\circ ,$ 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 $180^\circ ,$ 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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