Question

If the non-parallel sides of a trapezium are equal, prove that the trapezium is cyclic.

Hint: For a cyclic quadrilateral, the sum of the pair of opposite angles is ${180}^{\circ }$. Use this result to prove the trapezium is cyclic.

Answer 1

Consider trapezium ABCD such that AB is parallel to CD and $AD=BC$.
Consider two points M and N that can be considered as the foot of perpendicular drawn on AB from vertices D and C respectively.
Now, if we compare $\mathrm{\Delta }DAM$and $\mathrm{\Delta }CBN$, we have:
$AD=BC...\text{(Given)}$
$\mathrm{\angle }AMD=\mathrm{\angle }BNC...\text{(Right angles i}\text{.e}\text{. 9}{\text{0}}^{\circ }\text{)}$
And $DM=CN....\text{(Distance between two parallel lines)}$
From this we can say that both the triangles are congruent.
$\mathrm{\Delta }DAM\cong \mathrm{\Delta }CBN$
We know that corresponding parts of congruent triangles are equal. So we have:
$\Rightarrow \mathrm{\angle }A=\mathrm{\angle }B.....(i)$
Also $\mathrm{\angle }B+\mathrm{\angle }C={180}^{\circ }...\text{(sum of co - interior angles)}$
Substituting the value of $\mathrm{\angle }B$ from equation first, we have:
$\Rightarrow \mathrm{\angle }A+\mathrm{\angle }C={180}^{\circ }....(ii)$
Equation $(ii)$ shows that the sum of the pair of opposite angles of trapezium ABCD is ${180}^{\circ }$.
Therefore the trapezium is a cyclic quadrilateral.
Note: A quadrilateral is said to be cyclic quadrilateral if all of its 4 vertices lie on the same circle.