Question

Let $ABCD$ is a square with sides of unit length. Points $E$ and $F$ are on sides $AB$ and $AD$ respectively, so that $AE=AF.$ Let $P$ be a point inside square $ABCD.$
The value of $(PA{)}^2-(PB{)}^2+(PC{)}^2-(PD{)}^2$ is equal to

• A. $3$
• B. $2$
• C. $1$
• D. $0$

Answer 1

The correct option is D $0$

Refer the figure and apply Pythagoras Theorem:

$(PA{)}^2-(PD{)}^2=(PE{)}^2+(AE{)}^2-(PG{)}^2-(GD{)}^2=(PE{)}^2-(PG{)}^2....\left(1\right)$ and

$(PC{)}^2-(PB{)}^2=(PG{)}^2+(GC{)}^2-(PE{)}^2-(BE{)}^2=(PG{)}^2-(PE{)}^2....\left(2\right)$

Adding $\left(1\right)$ and $\left(2\right)$ gives the result immediately:

$(PA{)}^2-(PB{)}^2+(PC{)}^2-(PD{)}^2=0$

Thus, option D is the right answer