Question

in the given figure ABCD is a square and angle PQR is 90 if PB=QC=DR Prove that QB=RC

Answer 1

Step-by-step explanation:

It is given that O is a point within a quadrilateral ABCD Consider △ ACO We know that in a triangle the sum of any two sides is greater than the third side We get OA + OC > AC ….. (1)

In the same way Consider △ BOD We get OB + OD > BD ….. (2) By adding both the equations we get OA + OC + OB + OD > AC + BD Therefore, it is proved that OA + OC + OB + OD > AC + BD.

Answer 2

Explanation : We know that the line segment QB can be written as

QB = BC - QC

Since ABCD is a square, we know that BC = DC and QC = DR

So we get

QB = CD - DR

from the figure we get

QB = RC