Question

In a rhombus ABCD show that diagonal AC bisects $\angle A$ as well as $\angle C$ and diagonal BD bisects $\angle B$ as well as $\angle D$.

Answer 1

Given: A rhombus ABCD.
To prove:

Diagonal AC bisects $\angle A$ as well as $\angle C$ and diagonal BD bisects $\angle B$ as well as $\angle D$.
Proof:
In $△ABC$
AB = BC (Sides of rhombus are equal.)
∠4$\angle 4$ =$\angle 2$ (Angles opposite to equal sides are equal.)
...(1)
Now,
$AD\parallel BC$
(Opposite sides of rhombus are parallel.)
AC is transversal.
So, $\angle 1$ =$\angle 4$
(Alternate interior angles)
...(2)
From (1) and (2), we get
$\angle 1$ =$\angle 2$
Thus, AC bisects $\angle A$
Similarly,
Since, $AB\parallel DC$ and AC is transversal.
So, $\angle 2$ =$\angle 3$
(Alternate interior angles)
...(3)
From (1) and (3), we get
$\angle 4$ =$\angle 3$
Thus, AC bisects $\angle C$
Hence, AC bisects $\angle C$ and $\angle A$
In $△DAB$
AD = AB (Sides of rhombus are equal.)
$\angle ADB=\angle ABD$ (Angles opposite to equal sides are equal.)
Now,
$DC\parallel AB$
(Opposite sides of rhombus are parallel.)
BD is transversal.
So, $\angle CDB=\angle DBA$
(Alternate interior angles)...(5)...(4)

From (4) and (5), we get
$\angle ADB=\angle CDB$
Thus, DB bisects $\angle D$
Similarly,
Since, $AD\parallel BC$ and BD is transversal.
So, $\angle CBD$ =\angle ADB \) Alternate interior angles)...(6)
From (4) and (6), we get
$\angle CBD$ =\angle ADB \)
Thus, BD bisects \angle B \)
Hence, BD bisects $\angle D$ and $\angle B$