
$ABCD$ is a rhombus. Show that diagonal $AC$ bisects $\angle A$ as well as $\angle C$ and diagonal $BD$ bisects $\angle B$ as well as $\angle D$.
Answer
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Hint: All the sides of a rhombus are equal and the opposite sides are parallel to each other. Also, in a rhombus the angles opposite to equal sides are always equal.
Complete step by step solution:
The following is the schematic diagram of a rhombus.
Consider $\Delta ABC$,
Since all the sides of rhombus are equal, therefore
$AB = BC$
The angles opposite to the sides $AB$ and $BC$ will be equal. hence
$\angle 4 = \angle 2$.….(i)
Also $AD\parallel BC$ with transversal $AC$, as $AD$ and $BC$ are opposite sides of rhombus and opposite sides of rhombus are parallel to each other. Hence alternate angles will be equal.
$\angle 1 = \angle 4$……(ii)
From equation (i) and (ii).
$\angle 1 = \angle 2$
Hence, it is clear that $AC$ bisects the angle $\angle A$.
Now $AB\parallel DC$ with transversal $AC$, as $AB$ and $DC$ are opposite sides of rhombus and opposite sides of rhombus are parallel to each other. Hence alternate angles will be equal.
$\angle 2 = \angle 3$……(iii)
From equation (i) and (iii).
$\angle 4 = \angle 3$
Hence, it is clear that $AC$ bisects the angle $\angle C$.
Therefore $AC$ bisects angles $\angle A$ and $\angle C$.
Since $CD$ and $BC$ are the sides of rhombus, therefore $CD = BC$.
The following is the schematic diagram of a rhombus.
The angles opposite to the sides $CD$ and $BC$ will be equal. hence
$\angle 5 = \angle 7$..….(iv)
Also $AB\parallel CD$ with transversal $BD$ , as $AB$ and $CD$ are opposite sides of rhombus and opposite sides of rhombus are parallel to each other. Hence alternate angles will be equal.
$\angle 5 = \angle 8$……(v)
From equation (iv) and (v).
$\angle 7 = \angle 8$
Hence $BD$ bisects the angle $\angle B$.
Now $AD\parallel BC$ with transversal $BD$, as $AD$ and $BC$ are opposite sides of rhombus and opposite sides of rhombus are parallel to each other. Hence alternate angles will be equal.
$\angle 6 = \angle 7$……(vi)
From equation (v) and (vi).
$\angle 5 = \angle 6$
Hence $BD$ bisects the angle $\angle D$.
Therefore $BD$ bisects angles $\angle B$ and $\angle D$.
Note: Angle bisector divides the angle in two equal angles. Make sure to use the properties of rhombus in the solution.
Complete step by step solution:
The following is the schematic diagram of a rhombus.
Consider $\Delta ABC$,
Since all the sides of rhombus are equal, therefore
$AB = BC$
The angles opposite to the sides $AB$ and $BC$ will be equal. hence
$\angle 4 = \angle 2$.….(i)
Also $AD\parallel BC$ with transversal $AC$, as $AD$ and $BC$ are opposite sides of rhombus and opposite sides of rhombus are parallel to each other. Hence alternate angles will be equal.
$\angle 1 = \angle 4$……(ii)
From equation (i) and (ii).
$\angle 1 = \angle 2$
Hence, it is clear that $AC$ bisects the angle $\angle A$.
Now $AB\parallel DC$ with transversal $AC$, as $AB$ and $DC$ are opposite sides of rhombus and opposite sides of rhombus are parallel to each other. Hence alternate angles will be equal.
$\angle 2 = \angle 3$……(iii)
From equation (i) and (iii).
$\angle 4 = \angle 3$
Hence, it is clear that $AC$ bisects the angle $\angle C$.
Therefore $AC$ bisects angles $\angle A$ and $\angle C$.
Since $CD$ and $BC$ are the sides of rhombus, therefore $CD = BC$.
The following is the schematic diagram of a rhombus.
The angles opposite to the sides $CD$ and $BC$ will be equal. hence
$\angle 5 = \angle 7$..….(iv)
Also $AB\parallel CD$ with transversal $BD$ , as $AB$ and $CD$ are opposite sides of rhombus and opposite sides of rhombus are parallel to each other. Hence alternate angles will be equal.
$\angle 5 = \angle 8$……(v)
From equation (iv) and (v).
$\angle 7 = \angle 8$
Hence $BD$ bisects the angle $\angle B$.
Now $AD\parallel BC$ with transversal $BD$, as $AD$ and $BC$ are opposite sides of rhombus and opposite sides of rhombus are parallel to each other. Hence alternate angles will be equal.
$\angle 6 = \angle 7$……(vi)
From equation (v) and (vi).
$\angle 5 = \angle 6$
Hence $BD$ bisects the angle $\angle D$.
Therefore $BD$ bisects angles $\angle B$ and $\angle D$.
Note: Angle bisector divides the angle in two equal angles. Make sure to use the properties of rhombus in the solution.
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