0
$\begingroup$

Point $C$ lies inside a given right angle and points $A$and$B$ lie on its sides . Prove that the perimeter of $\Delta ABC$ is grater than $2OC$ , where $O$ is the vertex of the right angle . The figure is shown below:enter image description here

The solution given in the book is as follows :

We can reflect the points $C$ in the line $OA$ and $OB$ , to obtain the point $C'$ and $C''$, it is easy to see that point $O$ lies on the straight line $C'C''$ .Then we can replace perimeter of $\Delta ABC$ with the sum of the line segments $C'A,AB$ and $BC''$ . The triangle inequality tells us that this sum is no less than the length of $C'C''$ .This in turn is equal to $2OC$ ,since it is the hypotenuse of a right angle triangle of which $OC$ is the median .

However, I did not get the part where it says"that point $O$ lies on the straight line $C'C''$" ...I mean how are they drawing this conclusion. I am not getting ..also how are they drawing the conclusion that "This in turn is equal to $2OC$ "...I am not getting the idea behind this solution....

$\endgroup$

2 Answers 2

0
$\begingroup$

If $C = (x, y)$ then $C' = (-x, y)$ and $C'' = (x, -y)$, so $C'C''$ pass through the point $O$. Since $OC = OC'$ and $OC = OC''$, so $C'C'' = C'O + OC'' = 2OC$.

$\endgroup$
0
$\begingroup$

You are reflecting point $C$ about $OA$ and $OB$. See the below diagram with angles marked. It should become clear why $C'C"$ is a straight line and equal to $2 OC$.

enter image description here

$\endgroup$
2
  • $\begingroup$ I got that part .. i mean how $C'C''=2OC$ by completing the rectangle $C,C',C''$ but can u please explain why $C'C''$passes through $O$.... $\endgroup$
    – user992622
    Apr 16, 2022 at 14:01
  • 1
    $\begingroup$ Assume $\angle AOC = \alpha$ then $\angle BOC = 90 - \alpha$. Now reflection around $OA$ and $OB$ would make the same angles $\alpha$ and $90-\alpha$. So adding the angles, you get $180^\circ$ $\endgroup$
    – Math Lover
    Apr 16, 2022 at 14:12

You must log in to answer this question.