Question

Diagonal $AC$ of a parallelogram $ABCD$ bisects $\angle$ $A$. Show that
(i) it bisects $\angle$ $C$ also,
(ii) $ABCD$ is a rhombus.

Answer 1

diagonal $AC$ of $||gmABCD$ bisects $\angle A$

Also given, to show that, it bisects $\angle C$

Let us first know, what is a parallelogram "A quadrilant in which both pairs of opposite sides are parallel is called a parallelogram"

And theoretically Rhombus is a "Quadrilateral in which all four sides are equal to both pairs of opposite sides are parallel is called rhombus".

as given: $\overline{AC}$ diagonal bisects $\angle A$ in $ABCD||$ & $\angle CAD=\angle CAB$ (as given)

required to prove: (i) it bisects $\angle C$.

let us now consider.

proof: (i) In $\Delta AOC$ & $\Delta CBA$

$AD=CB$ ($\because$ common & opposite sides of $||gm$)

$DC=BA$ ($\because$ opposite sides of $||gm$)

$AC=CA$ ($\because$ common side)

$\therefore \Delta ADC\cong \Delta CBA$ [$\because$ by SSS congruency rule]

$\angle ADC=\angle CBA$

$\Rightarrow \angle ACD=\angle CAB$ ...(1) ($\because$ common angle)

$\angle BCA=\angle CAD$ ...(2) ($\because$ common angle)

$\angle CAB=\angle CAD$ ...(3) ($\because$ common angle)

from eq. (1), (2) & (3) we get,

$\angle ACD=\angle BCA$

$\therefore AC$ diagonal bisects $\angle C$.

Hence proved