Squared Parallelogram

Set A at origin 0 and B at z and C at w where z and w are complex. Remembering that multiplying by i performs a 90 degree rotation, then we identify the further vertices with complex numbers:

• I iz
• H z+iz
• G z+i(w-z)
• X z+iw

For the second parallelogram:

• E w-iw
• Y w-iz

To finish, Angle XAY = arg(z+iw)-arg(w-iz) = arg((z+iw)/(w-iz)) = arg(i) = 90 degrees.

Solution:

Denote R(O,ß) be the rotation with center O and the rotary angle ß. Writing +ß or ß means we rotate the figure clockwise. Now we will solve the problem in general position.

By R(A, $90 ^ \circ$ ), we can show that CI is perpendicular to BD (1).

In an other hand, we see that: EC is perpendicular to AC, EY is perpendicular to CB. Hence $\angle ACB = \angle CEY$ . Having that, we easily prove triangle CEY is equal to triangle ACB. Then: (2) CY = AI (3) $\angle ECY= \angle CAB$ , so CY is perpendicular to AB (since CE is perpendicular to AC). It shows that CY is parallel to AI.

Combining (2) and (3), we have that CYAI is a parallelogram, so AY is parallel to CI (4).

By the same way, we obtain AX is parallel to BD (5)

From (1), (4) and (5), it point out the angle XAY is equal to 90 degree. The problem is solved.

Since XBH is congruent to CAB by SAS (XH=BG=BC, BH=AB, <BHX=180-<HBG=<ABC), and BH is perpendicular to AB and XH (extended), which is parallel to BG, is perpendicular to BC, XB must be perpendicular to its corresponding side, AC. But AD is also perpendicular to AC and has equal length. Thus, we can translate XA to BD without rotation, preserving its angle with AY. We can similarly translate AY to IC. We thus only need to find the angle between BD and IC (facing C and D) Now, consider the rotation centered at A which rotates B into I. This obviously is a 90 degree rotation, given that <BAI=90. But <DAC is also 90 degrees, so the rotation will also rotate D into C. Thus BD and IC are perpendicular, so the angle between BD and IC is 90 degrees in any direction. So the answer must be 90 degrees

We know $XG=HB=AB=23$ and $BG=BC=32.$ Furthermore, $\angle XGB = 180^\circ-\angle \angle HBG = \angle ABC.$ It follows that from SAS congruence, $\triangle XGB \cong\triangle ABC$ . SImilarly, $\triangle YCF \cong\triangle ABC.$ Consequently, $\triangle ACY \cong\triangle XBA$ also from SAS.

Now, let $\angle BAC = \alpha, \angle ABC = \beta, \angle ACB= \gamma.$ The desired $\angle XAY$ is equal to $\alpha+\angle XAB+\angle YAC = \alpha + 180^\circ - \angle ACY = \alpha + 180^\circ - (360^\circ- \gamma - 90^\circ - \beta).$

This expression simplifies to $\alpha+\beta+\gamma-90^\circ = 90^\circ,$ which is our answer.

Note:

$\angle ABC = 180^\circ - \angle GBH = \angle XGB$

$BA = BH = GX$

$BC = GB$

By SSA, we have $\triangle ABC \equiv \triangle XGB$

By similar arguments, we have:

$\triangle ABC \equiv \triangle CYE \equiv BHX$

Note that $\angle XBA = 90^\circ+\angle XBH = 90^\circ+\angle ECY = \angle ACY$ .

It is easy to see that $AB=YC$ and $XB=AC$ .

Thus we have $\triangle CAY \equiv \triangle BCA$

We thus have $\angle BAX+\angle CAY+\angle XBA=180^\circ$ .

Note that $\angle XBA = \angle XBH + 90^\circ = \angle CAB + 90^\circ$

Hence $\angle XAY = \angle BAX+\angle CAY+\angle CAB$

$= \angle BAX+\angle CAY+(\angle XBA - 90^\circ) = 180^\circ-90^\circ=90^\circ$

Note that $\angle{BGC}=180-\angle{HBG}=\angle{ABC}$ , $BG=BC$ , and $XG=BH=AB$ , so $\triangle{XGB}\cong\triangle{ABC}$ by SAS. By similar reasoning, $\triangle{YCF}\cong\triangle{ABC}$ . It follows that

$\angle{ABX}=360-\angle{ABH}-\angle{ABC}-\angle{XBG}$ $=360-90-\angle{ABC}-\angle{ACB}=270-(180-\angle{BAC})=90+\angle{BAC}$

and

$\angle{ACY}=360-\angle{ACE}-\angle{ACB}-\angle{YCF}$ $=360-90-\angle{ACB}-\angle{ABC}=270-(180-\angle{BAC})=90+\angle{BAC}$ ,

so $\angle{ABX}=\angle{ACY}$ . But $AB=YC$ and $AC=XB$ , so $\triangle{ABX}\cong\triangle{YCA}$ by SAS. Hence,

$\angle{XAY}=\angle{XAB}+\angle{CAY}+\angle{BAC}=\angle{XAB}+\angle{BXA}+\angle{BAC}$ $=180-\angle{ABX}+\angle{BAC}=180-(90+\angle{BAC})+\angle{BAC}=\boxed{90}$ .

$\angle GBH = 180^ \circ - \angle B$ $\angle BGX = \angle B$ Triangles ABC , XGB and YCF are congruent triangles. Also triangles ABX and YCA are also congruent. Therefore, $\angle XAY = \angle XAB + \angle BAC + \angle CAY$ $= \left( {\angle XAB + \angle BXA} \right) + \angle BAC$ $= \left( {180^ \circ - \left( {90^ \circ + \angle BAC} \right)} \right) + \angle BAC$ $= 90^ \circ$

xd straight line and iy are stright lines hence angle xay = 90

AB=BH=XG=23

AC=CE=FY=44

angle HBG= 360 - (90+90+B) = 180-B and angle ECF is similarly 180-C.

Thus angle CFY=C and BGX=B.

Thus, XGB is congruent to ABC (BG=BC=32, AB=XG=23 and the containing angle B). Thus BX=44 and angle BXG=A

Similarly, CFY is congruent to ABC. Thus CY=23 and angle CYF=A

Now triangle ABX is congruent to YCA (since AB=YC=23, BX=AC=44 and the containing angle 90+A)

Thus angle XAC=AYC=x(say) and angle AXC=CAY=y(say). So, x+y= 180 - (90 + A)=90 - A

Thus the required angle = 90- A +A = 90