Question
Give a geometrical construction for finding the fourth point lying on a circle passing through three given points, without finding the centre of the circle. Justify the construction.
✓
Answer
Let $A, B$ and $C$ be the given points. With $B$ as the centre and a radius equal to $AC$, draw an arc. With $C$ as the centre and $AB$ as radius, draw another arc, which cuts the previous arcat $D$.
Then $D$ is the required point $BD$ and $CD$.
In $\triangle\text{ABC}$ and $\triangle\text{DCB}$
$AB = DC\ AC = DB\ BC = CB$ [Common]
$\therefore\ \triangle\text{ABC}\cong\triangle\text{DCB}$ [By $SSS$]
$\Rightarrow\ \angle\text{BAC}=\angle\text{CDB}$ $[C.P.C.T.]$
Thus, BC subtends equal angles, $\angle\text{BAC}$ and $\angle\text{CDB}$ on the same side of it.
$\therefore$ Points $A, B, C, D$ are concyclic.
Then $D$ is the required point $BD$ and $CD$.
In $\triangle\text{ABC}$ and $\triangle\text{DCB}$
$AB = DC\ AC = DB\ BC = CB$ [Common]
$\therefore\ \triangle\text{ABC}\cong\triangle\text{DCB}$ [By $SSS$]
$\Rightarrow\ \angle\text{BAC}=\angle\text{CDB}$ $[C.P.C.T.]$
Thus, BC subtends equal angles, $\angle\text{BAC}$ and $\angle\text{CDB}$ on the same side of it.
$\therefore$ Points $A, B, C, D$ are concyclic.
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