# What is the measurement of the angle x in the figure below?

For reference: In the figure, ABCD is a parallelogram and $$\measuredangle ABE = 100^o$$ , $$AE = AD$$ and $$FC = CD$$. Calculate x. My progress:

In this one I couldn't develop much because $$FC$$ and $$EB$$ are not parallel...I think the "output" is by some auxiliary construction by geogebra • You say that $\angle ABE=145$, but your diagram suggests that it is $100$. Is there a typo? Aug 20, 2021 at 14:33
• $\Delta EAB \cong \Delta BCF$ by SAS congruency criterion Aug 20, 2021 at 14:50
• @peta arantes, in figure $\angle ABE=100 ^o$ which one is correct? Aug 20, 2021 at 16:35
• @AlanAbraham..my mistake, thanks for the alert..I've corrected the statement Aug 20, 2021 at 18:52
• @sirous my mistake, thanks for the alert..I've corrected the statement Aug 20, 2021 at 18:52

In $$\Delta EAB$$ and $$\Delta BCF$$ \begin{align} EA&=BC &(\because AD=BC, \text{ opposite sides of parallelogram})\\ AB&=CF & (\because CD=AB, \text{ opposite sides of parallelogram)}\\ \angle EAB&=\angle BCF=y \text{ (let)} & (\because 90°-\angle BAC=90°-\angle BCD)\\ \therefore \Delta EAB &\cong \Delta BCF &\text{ (by SAS congruency criterion)}\\ \implies \angle EBA&=\angle BFC= 100°&\text{ (corresponding angles of congruent triangles)}\\ \text{and}\qquad EB&=BF &\text{ (corresponding sides of congruent triangles)}\\ \end{align}

In parallelogram $$ABCD$$,

\begin{align}&\angle BAD =90°-y\\ \implies &\angle ABC=180°-(90°-y)=90°+y \end{align}

In triangle $$BCF$$, $$\angle FBC= 180°-(100°+y)=80°-y$$

In triangle $$EBF$$, $$\angle BEF=\angle BFE=x-100° \; (\because EB=BF)$$

$$\implies \angle EBF=180°-2(x-100°)=380°-2x$$

Sum of all angles around point $$B$$ $$= 100°+(90°+y)+(80°-y)+(380°-2x)=360°$$

$$\therefore x=145°$$

• $\angle EBA= 145°$ is not possible Aug 20, 2021 at 15:19
• @Aman..great resolution.. grateful Aug 20, 2021 at 18:54
• @petaarantes you are welcome Aug 21, 2021 at 1:44 All you need is to show points F, B, D and C are cyclic, then we have:

$$\angle ADB=45$$

$$\angle FBD=90$$

$$\rightarrow \angle FAB=35\rightarrow \angle BAD=\angle BCD=55=\angle BFD$$

$$\Rightarrow \angle ABD=80$$

$$\angle BFC=55+45=100$$

$$\angle BEF=45$$

$$\angle FBE=360-(100+80+90)=90$$

$$\Rightarrow \angle EFB=45$$

Finally:

$$x=45+100=145$$

• @sirous...EB is not parallel to FC so I don't think it's cyclic..see my last figute..$\rightarrow \angle FAB=35$ I don't understand Aug 20, 2021 at 19:17

Another solution. Perform a $$90^{\circ}$$ clock-wise rotation around $$D$$ and let the rotated image of $$A$$ be denoted by $$H$$. Then $$AD = DH = AE$$ which means that $$ADHE$$ is a square. If you rotate $$90^{\circ}$$ counter-clock-wise around $$D$$ the segment $$CD$$, you get the segment $$GD$$ such that $$GD = CD$$ and $$GD \, \perp \, CD$$. Hence, $$CDGF$$ is a square. Because of these new extra constructions, during the $$90^{\circ}$$ clock-wise rotation around $$D$$, The points $$A$$ and $$G$$ are rotated to the points $$H$$ and $$C$$ and consequently, the triangle $$ADG$$ is rotated to the triangle $$HDC$$ which means that $$\angle\, HCD = \angle\, AGD$$ $$HC = AG$$ $$HC \, \perp \, AG$$ Now, if you translate triangle $$HDC$$ along $$DA$$ you get as a parallel translated image the triangle $$EAB$$. Furthermore, if you translate triangle $$ADG$$ along $$AB$$ you get as a parallel translated image the triangle $$BCF$$. Therefore, $$\angle\, EBA = \angle \, BFC = 100^{\circ}$$ $$EB = FB$$ $$EB \, \perp \, FB$$ Hence, triangle $$EBF$$ is right-angled and isosceles, which means that $$\angle \, EFB = 45^{\circ}$$ Now putting all these angles together $$x = \angle \, EFC = \angle\, EFB + \angle \, BFC = 45^{\circ} + 100^{\circ} = 145^{\circ}$$