Question

Two sides DA and BC of quadrilateral ABCD are produced as shown in the given figure. Show that \[m + n = x + y\]

Answer 1

Hint: To solve this given problem we will use the properties of a quadrilateral along with the properties of the linear paired angles. We can write angles \[\angle DAB\,\,\& \,\,\angle DCB\] in the forms of m and n. Now as the sum of all the angles is 360 degrees, putting those values in the right places will give us the answer.

Complete step by step answer:

According to the given figure we get,
As, we have a linear pair at angle A, we get,
\[\angle DAB + m^\circ = 180^\circ \;\]
\[ \Rightarrow \angle DAB = 180^\circ - m^\circ .....\left( 1 \right)\]
Again, at angle C we have a linear pair, so we get,
\[\angle DCB + n^\circ = 180^\circ \]
\[ \Rightarrow \angle DCB = 180^\circ - n^\circ .....\left( 2 \right)\]
Also, as the Sum of angles of a quadrilateral is \[360^\circ \]
\[ \Rightarrow \]\[\angle DCB + \angle DAB + x^\circ + y^\circ = 360^\circ \;{\text{ }}\;\]
Putting the values of equation 1 and equation 2, we get,
\[ \Rightarrow 180^\circ - m^\circ + 180^\circ - n^\circ + + x^\circ + y^\circ = 360^\circ \;{\text{ }}\;\]
On adding like terms we get,
\[ \Rightarrow 360^\circ - m^\circ - n^\circ + + x^\circ + y^\circ = 360^\circ \;{\text{ }}\;\]
On cancelling out common values we get,
\[ \Rightarrow - m^\circ - n^\circ + x^\circ + y^\circ = 0{\text{ }}\;\]
On rearranging we get,
\[ \Rightarrow x^\circ + y^\circ = m^\circ + n^\circ {\text{ }}\;\]
Hence proved.

Note: There are two properties of quadrilaterals we should also know,
1) A quadrilateral should be closed shape with 4 sides
2) All the internal angles of a quadrilateral sum up to\[360^\circ \]