Question

ABCD is a quadrilateral, diagonal AC and BD bisect each other at right angles. Prove that ABCD is a rhombus​

Answer 1

Answer:

we have now that the properties of rhombus

1st =opposite sides are congruent.

2nd =diagonals of rhombus are perpendicular bisector of each other.

and similarly,

we can prove that,

the quadrilateral ABCD is a rhombus.

because , in the second property means in the property of rhombus we already studied that, diagonals of rhombus are perpendicular bisector of each other.

I think the answer may be helpful for you!!

Answer 2

Answer:

$\setlength{\unitlength}{1cm}\begin{picture}(0,0)\thicklines\put(0,0){\line(1,3){1.5}}\put(0,0){\line(1,0){5}}\put(5,0){\line(1,3){1.5}}\put(1.5,4.5){\line(1,0){5}}\qbezier(1.56,4.5)(1.56,4.5)(5,0)\qbezier(6.45,4.5)(6.45,4.5)(0,0)\put(-0.5,-0.5){\sf B}\put(1,4.8){\sf A}\put(5.2,-0.5){\sf C}\put(6.7,4.75){\sf D}\put(3,1.6){\sf O}\end{picture}$

Here, we have a quadrilateral ABCD, with AC and BD bisecting each other at right angles at point O.

From this quadrilateral, we have to prove that it is a rhombus by proving the congruency of any two triangles.

Taking ΔBOC and ΔBOA,

BO = BO [Common]

OA = OC [O is the midpoint of AC]

∠BOC = ∠BOA [Right angles]

∴ ΔBOC ≌ ΔAOD by SAS criterion.

So, by taking CPCT,

BC = BA ------------- (1)

BC = CD ------------ (2)

BA = AD ------------ (3)

AD = CD ------------ (4)

From (1), (2), (3) and (4), we can infer that BC = CD and BA = AD.

∴ Quadrilateral ABCD is a rhombus.

Knowledge Bytes:

→ Congruency rule - SAS

SAS ( Side-Angle-Side) congruency rule is used in those triangles where 2 sides and included angle of one triangle is congruent to the 2 sides and the included side of the other triangle.

→ CPCT

When two triangles are proved congruent by congruency rules, their corresponding parts are also congruent by CPCT (Corresponding parts of congruent triangles).

→ Rhombus

A parallelogram with the 4 sides equal is known as a rhombus. Its diagonals bisect each other at right angles.