Rotating a Triangle

Since the triangle $A'B'C$ is formed by rotating $ABC$ 90 degree counter-clockwise acbout $C$ , so the two triangles $A'B'C$ and $ABC$ are equal. Thus $B'C=BC$ and $A'C=AC$ .

Let $E$ be the point on $CD$ extended such that $ED=CD$ . We have $CE$ and $B'A$ bisect each other at $D$ , so $ACB'E$ is a parallelogram. This implies $AE=B'C=BC$ ; and $\angle EAC+\angle B'CA=180^\circ$ . On the other hand, $180^\circ=90^\circ+90^\circ=\angle B'CB +\angle A'CA= \angle B'CA +\angle BCA'$ . Hence, $\angle EAC=\angle BCA'$ . Together with the conclusions above that $AC=A'C$ and $AE=BC$ , so the two triangles $A'CB$ and $CAE$ are equal by S.A.S. Thus, $A'B=CE=2CD$ , which gives the answer $|CD|=601$ .

Let C be the origin and A and B are on the 2nd quadrant.

$A = z_1$ and $B = z_2$ and $C = 0$

Rotating counter-clockwise by C is equivalent of multiplying A and B by $i = e^{i\pi/2}$

So, $A'=i*z_1$ $B'=i*z_2$

$D=(z_1 + i*z_2)/2$

$\Rightarrow |CD|=|(z_1 + i*z_2)/2|$

$= |i|*|(z_1 + i*z_2)/2|$

$= |(i*z_1 - z_2)/2|$

$= |A'B/2|$

$= 1202/2$

$= 601$

Solution: $|CD|= 601$

Let the point C be at origin $(0,0)$ , the other points be at $A(x, y)$ and $B(z, 0)$ . Rotating 90 degree counter-clockwise about C, we get A'B'C. The coordinate of A' will be $(y, -x)$ , as CA' is perpendicular to CA. Similarly the coordinate of B' will be $(0, -z)$ . The coordinate of D is $\frac{x}{2}, \frac{y-z}{2}$ . We are given, $\sqrt{(y-z)^2 + x^2} = 1202$ Hence, $CD = \sqrt{(y-z)^2 + x^2} / 2 = 601$

First, rotate triangle $AB'C$ $90^\circ$ counter-clockwise about $C$ to form $A''B''C$ . $D$ is then mapped to $D'$ . Note that $A''=A$ (obvious), and that $BCB''$ is collinear (since $\angle BCB'= \angle B'CB''=90^\circ$ ). Now we compare triangles $B''CD'$ and $B''BA'$ . Since $|B''C| = |CB|$ and $|B''D'| = |D'A'|$ , we know that $\frac{|B''C|}{|B''B|} = \frac{|B''D|}{|B''A'|} = \frac{1}{2}$ . Also both triangles share a common angle $BB''A$ , so they're similar. Therefore, $\frac{|CD'|}{|BA'|} = \frac{1}{2}$ . Hence $|CD| = |CD'| = \frac{1}{2} \times 1202 = 601$ .

let B(a,o) >let A(x,y) then B'(0,a) >then A'(-y,x) from distance formula:- (x x +(a+y) (a+y))=(1202) (1202) -------------(1) (OC) (OC)=((x/2) (x/2)+(a/2+y/2) (a/2+y/2)) FROM ----(1) OC =1202/2=601//

Extend $CD$ to a point $E$ such that $CD=DE$ . Also, it is given that $AD=DB$ and by opposite angles, $\angle ADC = \angle EDB'$ . Thus, triangles $EDB'$ and $CDA$ are congruent. Also, $\angle DEB' = \angle DCA$ , which implies that $EB' \parallel AC$ . By a similar argument, triangles $CDB'$ and $EDA$ are congruent, which shows $EA \parallel B'C$ . Therefore, $AEB’C$ is a parallelogram. Since $AE = CB’=CB$ , $AC = CA’$ , $\angle CAE = 180^\circ - \angle ACB’ = 90^\circ + (90 - \angle ACB') = 90^\circ + \angle ACB = \angle BCA’$ , hence triangles $ACE$ and $CA’B$ are congruent. Thus, $CD = \frac {1}{2} CE = \frac {1}{2} A’B = 601$ .

Note: It doesn't matter if $ABC$ is labeled clockwise or anti-clockwise. Similarily, it doesn't matter if the rotation is clockwise or anti-clockwise. A direction is specified to avoid ambiguity in the above solution.