Question

Prove that in a parallelogram, the bisector of any two consecutive angles intersect at the right angle.

Answer 1

Hint: Parallelogram is a simple quadrilateral with two pairs of parallel sides. The opposite sides of a parallelogram are equal in length, and also opposite angles of a parallelogram are equal. One of the properties of a parallelogram says that the sum of any two consecutive angles is 180; hence in this question, find the value of the two consecutive angles and then find their bisecting angles.
In this question, we need to prove that the bisector of any two consecutive angles intersect at the right angle for which we need to follow the above discussed property of the parallelogram.

Complete step by step answer:
 In the given figure
Side AD is parallel to BC, and AB is parallel to DC where the side BC is transversal, hence
$$\begin{gathered} AD\parallel BC \\ AB\parallel DC \end{gathered}$$

Since one of the properties of parallelogram say the sum of the consecutive angle of a parallelogram is a supplementary angle hence, we can write,
$$\mathrm{\angle }ABC+\mathrm{\angle }DCB={180}^{\circ }$$
Now draw a line OB bisecting $$\mathrm{\angle }B$$and a line OC bisecting $$\mathrm{\angle }C$$joining at O since line OB and OC bisects the angle; hence we can say
$$\mathrm{\angle }OBC={\displaystyle \frac{\mathrm{\angle }ABC}{2}}$$
$$\mathrm{\angle }OCB={\displaystyle \frac{\mathrm{\angle }DCB}{2}}$$
Hence in the $$△OBC$$
$$\begin{gathered} \mathrm{\angle }OBC+\mathrm{\angle }OCB+\mathrm{\angle }BOC=180 \\ x+y+\mathrm{\angle }BOC=180 \end{gathered}$$
This can be written as
$${\displaystyle \frac{\mathrm{\angle }ABC}{2}}+{\displaystyle \frac{\mathrm{\angle }DCB}{2}}+\mathrm{\angle }BOC=180$$
Hence by solving
$${\displaystyle \frac{1}{2}}\left(\mathrm{\angle }ABC+\mathrm{\angle }DCB\right)+\mathrm{\angle }BOC=180$$
Since$$\mathrm{\angle }ABC+\mathrm{\angle }DCB={180}^{\circ }$$hence we can write

$$\begin{gathered} {\displaystyle \frac{1}{2}}\left(\mathrm{\angle }ABC+\mathrm{\angle }DCB\right)+\mathrm{\angle }BOC=180 \\ {\displaystyle \frac{1}{2}}\times 180+\mathrm{\angle }BOC=180 \\ 90+\mathrm{\angle }BOC=180 \\ \mathrm{\angle }BOC={90}^{\circ } \end{gathered}$$
So the value of$$\mathrm{\angle }BOC={90}^{\circ }$$,
Therefore we can say the bisector of any two consecutive angles intersect at the right angle.
Hence proved.

Note: A property of that parallelogram says that if a parallelogram has all sides equal, then their diagonal bisector intersects perpendicularly. An angle bisector is a ray that splits an angle into two consecutive congruent, smaller angles.