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Question

Show that if the diagonals of a quadrilateral are equal and bisect each other at right angles, then it is a square.


Solution

Given: 
$$\Box ABCD$$ is a quadrilateral.
diag $$AC =$$ diag $$BD$$, intersecting at $$E$$.
$$AC$$ and $$BD$$ are perpendicular bisectors of each other
$$\therefore \angle E=90^o$$

To prove: 
$$\Box ABCD$$ is a square.

Solution: 
In $$\triangle ABE$$ and $$\triangle ADE$$.
$$BE = DE$$              ....given
$$AE = AE$$              ...common side
$$\angle AEB \cong \angle AED$$     ....each $$90^o$$
$$\therefore \triangle ABE \cong \triangle ADE$$     ...SAS test of congruence
$$\therefore AB = AD$$                  ...c.s.c.t.      ....(1)

Similarly, we can prove $$\triangle ABE \cong \triangle CBE$$
$$\therefore AB = CB$$    ....c.s.c.t.        ....(2)
And, $$\triangle ADE \cong \triangle CDE$$
$$\therefore AD = CD$$    ....c.s.c.t.        ....(3)

$$\therefore$$ From (1), (3) and (4),
$$ AB = CB = CD = AD$$     ............(4)

Thus, in quad.ABCD
$$ AB = CB = CD = AD$$  and   $$AC=BD$$ [given]
$$\therefore \Box ABCD$$ is a square        ....By definition

496311_463878_ans_bfe59aaa071c4588a8b86f5608ef773a.png

Mathematics
RS Agarwal
Standard IX

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