Question

The diagonals of a rhombus bisect each other at
(a)$60{}^{\circ }$
(b)$80{}^{\circ }$
(c)$90{}^{\circ }$
(d)$120{}^{\circ }$

Answer 1

Hint: Rhombus is a quadrilateral with four equal sides whose diagonals bisect each other. Take any two adjacent triangles formed by the intersection of the diagonals and try to prove them congruent. Use the property that the measure of a straight angle is $180{}^{\circ }$, so a linear pair of angles must add up to $180{}^{\circ }$.

Complete step-by-step answer:
As we know the property of rhombus that all sides of the rhombus are equal to each other and diagonals bisect each other and here we need to determine the angle formed by the diagonals at the intersecting point.
So, let we have a rhombus ABCD which diagram can be given as :-

So, the sides AB, BC, CD and AD are equal to each other and diagonals AC and BD bisecting each other i.e. length AO is equal to CO and BO is equal to DO.
So, we have
$AB=BC=CD=AD$ …......................................(i)
$AO=CO$ …......................................(ii)
$BO=DO$ …...................................(iii)
So, from the given diagram, we have to prove angles $\mathrm{\angle }AOB,\mathrm{\angle }DOA,\mathrm{\angle }DOC,\mathrm{\angle }COB$ are $90{}^{\circ }$.
In $\mathrm{\Delta }AOB$ and $\mathrm{\Delta }BOC$, we have
$AO=CO$ (from equation (ii))
$AB=BC$ (from equation (i))
$BO=BO$ (common in both triangles).
Hence, $\mathrm{\Delta }AOB$ and $\mathrm{\Delta }BOC$ are congruent to each other by SSS criteria of congruence. So, we get
$\mathrm{\angle }AOB=\mathrm{\angle }BOC$ (By C.P.C.T.) ………………………………….(iv)
Now, we know that the sum of angles formed by a line on a line is $180{}^{\circ }$ because of linear pair property.
It means the angles in the diagram below sum as $180{}^{\circ }$.

So, $\mathrm{\angle }1+\mathrm{\angle }2=180{}^{\circ }$ (Linear pair)
  Hence, from the rhombus, the sum of angles $\mathrm{\angle }AOB$ and $\mathrm{\angle }BOC$ is 180 by the above mentioned property. So, we get
$\mathrm{\angle }AOB+\mathrm{\angle }BOC=180{}^{\circ }$ ……………………………………………………(v)
Now, from the equation (iv), we have
$\mathrm{\angle }AOB=\mathrm{\angle }BOC$
So, we can re-write equation (v) as
$\mathrm{\angle }AOB+\mathrm{\angle }AOB=180{}^{\circ }$
$2\mathrm{\angle }AOB=180{}^{\circ }$
$\mathrm{\angle }AOB={\displaystyle \frac{180{}^{\circ }}{2}}=90{}^{\circ }$
Hence, $\mathrm{\angle }AOB=90{}^{\circ }$ and $\mathrm{\angle }BOC=90{}^{\circ }$ as well from the equation (iv).
Now, we know pairs of vertically opposite angles are always equal formed by intersection of two lines.
It means, we get from the rhombus ABCD as
$\mathrm{\angle }AOB=\mathrm{\angle }COD$
$\mathrm{\angle }BOC=\mathrm{\angle }AOD$
Hence, $\mathrm{\angle }COD=90{}^{\circ }$ and $\mathrm{\angle }AOD=90{}^{\circ }$ as well from the above conditions.
So, the diagonals of rhombus bisect each other at $90{}^{\circ }$.
Hence, option (c) is correct.

Note: One may take any two triangles from the pairs $(\mathrm{\Delta }AOB,\mathrm{\Delta }BOC)$, $(\mathrm{\Delta }BOC,\mathrm{\Delta }DOC)$, $(\mathrm{\Delta }AOD,\mathrm{\Delta }AOB)$, $(\mathrm{\Delta }DOC,\mathrm{\Delta }AOD)$ to prove the angle formed by diagonals as $90{}^{\circ }$. It is not necessary to take the triangles AOB and BOC as done in the solution.
Use the fundamental properties of a rhombus and don’t get confused with the properties of rectangle, parallelogram or square. All have some common properties and some different properties as well. So, be clear with the properties of a rhombus to solve these kinds of problems.