Question
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
✓
Answer
Given: $ABCD$ is a trapezium whose non-parallel sides $AD$ and $BC$ are equal.
To prove: Trapezium $ABCD$ is a cyclic Join $BE,$ where $BE || AD.$
Proof: Since, $AB || DE$ and $AD || BE$ Since, the quadrilateral $ABCD$ is a parallelogram, $\therefore\angle\text{BAD}=\angle\text{BED}\ \ ...(\text{i})$ [opposite angles of a parallelogram are equal] and $AD = BE ...(ii)$ [opposite sides of a parallelogram are equal] But,
$AD = BC [$given$] ...(iii)$ From Eqs. $(ii)$ and $(iii),$
$BC = BE$
$\Rightarrow\angle\text{BEC}=\angle\text{BCE}\ \ ...\text{(iv)}$ [angle opposite to equal sides are equal]
Also, $\angle\text{BEC}+\angle\text{BED}=180^\circ[$ linear pair axiom$]$
$\therefore\angle\text{BCE}+\angle\text{BAD}=180^\circ [$from Eqs. $(i)$ and $(iv)]$
If sum of opposite angles of a quadrilateral is $180^\circ $, then quadrilateral is cyclic.
Hence, trapezium $ABCD$ is a cyclic. Hence proved.
To prove: Trapezium $ABCD$ is a cyclic Join $BE,$ where $BE || AD.$
Proof: Since, $AB || DE$ and $AD || BE$ Since, the quadrilateral $ABCD$ is a parallelogram, $\therefore\angle\text{BAD}=\angle\text{BED}\ \ ...(\text{i})$ [opposite angles of a parallelogram are equal] and $AD = BE ...(ii)$ [opposite sides of a parallelogram are equal] But,
$AD = BC [$given$] ...(iii)$ From Eqs. $(ii)$ and $(iii),$
$BC = BE$
$\Rightarrow\angle\text{BEC}=\angle\text{BCE}\ \ ...\text{(iv)}$ [angle opposite to equal sides are equal]
Also, $\angle\text{BEC}+\angle\text{BED}=180^\circ[$ linear pair axiom$]$
$\therefore\angle\text{BCE}+\angle\text{BAD}=180^\circ [$from Eqs. $(i)$ and $(iv)]$
If sum of opposite angles of a quadrilateral is $180^\circ $, then quadrilateral is cyclic.
Hence, trapezium $ABCD$ is a cyclic. Hence proved.
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