# 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 $2OC$

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: 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....

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 $$C'C'' = C'O + OC'' = 2OC$$.
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 $$2 OC$$. • 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 $O$....
• 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 $180^\circ$ Apr 16, 2022 at 14:12