Question
If non-parallel sides of a trapezium are equal, prove that it is cyclic.
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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°, 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°, then quadrilateral is cyclic.
Hence, trapezium ABCD is a cyclic.
Hence proved.
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