How can I determine the value of angle x in this figure.
[![I used geometry but Did not find the answer. ][1]][1]
**I used Geometry but not avail ** [1]:
N:B The fiqure is a square.
Here is what I tred - When using regular geometry I'm getting 4 equations with 4 variables but these 4 just result in expressing the variables in the form of other variables. Equations:- X+Y=115, X+W=135, Y+Z=90, W+Z=110
3 Answers
$\begingroup$I've labelled a few points in the diagram; the angles in red should be trivial to find.
Now, make a copy of $\triangle ABE$, and rotate it about $A$ until side $AB$ coincides with side $AD$.
Now you just have to compare $ \triangle AE'F$ to $ \triangle AEF$. Do you see the rest?
$\endgroup$ $\begingroup$Draw a circle in center A and passes vertex E. This circle also psses D which is base of altitude from A. In this case triangles ADC and ACE are equal because AC is common side, AD and AE are radius of circle and CD=CE because they are tangents from C on the side BC. Therefore $\widehat {DAC}=\widehat{CAE}=25^o$ ,so $x=\widehat {ACD}=90-25=65^o$.
$\endgroup$ $\begingroup$Hint:
Fold $\triangle ADE$ inside along line $DE$ and $\triangle CDF$ inside along line $DF$.
As $\angle ADE + \angle CDF = 45^0 = \angle EDF, \angle A = \angle C = 90^0 $ and $AD = CD$.
Where will points $A$ and $D$ meet and what does that tell about length $EF$ and about $\angle DEF$ wrt. $\angle DEA$?
Please ignore rest of the points and dotted lines. Those can be used to come up with another proof though.
$\endgroup$ 4
