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.
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.
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