# Find x in below diagram geometry

I have drawn two parallel lines from D and E intersecting sides CB and CE respectively on F and G. look below I got x+y=70 and x+a+b=130. a=y, b=60. now how to proceed? Am I moving in right direction?

• Please explain what you've tried so far, your thoughts on how to proceed, etc. Commented Aug 21, 2015 at 12:38
• Done. Can you help now? Commented Aug 21, 2015 at 12:46
• Your attempt is a good thought, but I think you're going in the wrong direction. Let $P$ be the point of intersection of the line segments $AE$ and $BD$. Then can you find the angle $\angle APB$? Commented Aug 21, 2015 at 12:50
• APB=50 ADB=40 and AEB=30 Commented Aug 21, 2015 at 12:52
• This problem is actually harder than I thought at first.... I'm not sure what to do besides finding the coordinates of each point ($A, B, C, D, E, P$) and so on. Commented Aug 21, 2015 at 13:34

First: $ABC$ is a isosceles triangle and $\hat{C}=20^{\circ}$. This implies also that $\Delta CDB$ is isosceles. $AB = 2AC\sin(10^{\circ})$. Hence $$\frac{CE}{BE} = \frac{S_{ACE}}{S_{ABE}}=\frac{AC\sin(10^{\circ})}{AB\sin(70^{\circ})} = \frac{1}{2\sin(70^{\circ})} = \frac{\sin(30^{\circ})}{\sin(110^{\circ})}.$$ This also implies $$\frac{\sin(\hat{CDE})}{\sin(\hat{BDE})} = \frac{S_{CDE}}{S_{BDE}} = \frac{CE}{BE} = \frac{\sin(30^{\circ})}{\sin(110^{\circ})}.$$

Also $\hat{CDB} = 140^{\circ}$. From here I believe that $\hat{CDE} = 30^{\circ}$ which leads to $x = 20^{\circ}$.

• $\triangle$ACE and $\triangle$ABE are not isosceles. How you proved that? Commented Aug 21, 2015 at 15:10
• I haven't said that the two triangles are isosceles.
– M. T
Commented Aug 21, 2015 at 15:12
• i dont understand the whole line having $\frac{CE}{BE}$ Commented Aug 21, 2015 at 15:15
• Draw a perpendicular line from A to CB and you will see it. Basically, it is just two ways to calculate the ratio between the areas of these two triangles.
– M. T
Commented Aug 21, 2015 at 15:17
• $\frac{S_{ACE}}{S_{ABE}} ??$ Commented Aug 21, 2015 at 15:18

Another way to consider circumscribed circles and inscribed circles of triangles.

Let $H$ be the center of the circumscribed circle of $\triangle{AEB}$ and $I$ be the center of the inscribed circle of $\triangle{DAB}$.

• $H$ is on $DB$.

Take $J$ on $DB$ such that $\angle{JAB}=60^\circ$. Then, since $\triangle{JAB}$ is an equilateral triangle, $\angle{AJB}=60^\circ=2\times 30^\circ=2\angle{AEB}$. Hence, by the converse theorem on inscribed angles, $J=H$.

• $E,H,I$ are on a line.

Since $\triangle{EHB}$ is an isoscels triangle, $\angle{EHB}=180^\circ-\angle{HEB}-\angle{HBE}=180^\circ-20^\circ-20^\circ=140^\circ$. On the other hand, since $\angle{HIB}=\angle{IAB}+\angle{IBA}=80^\circ+30^\circ=110^\circ$, $\angle{IHB}=180^\circ-\angle{HIB}-\angle{HBI}=180^\circ-110^\circ-30^\circ=40^\circ$. Thus, $\angle{EHB}+\angle{IHB}=180^\circ$.

• $D,I,B,E$ are on a circle.

Since $\angle{IDB}=\frac 12\angle{ADB}=20^\circ$, $\angle{IDB}=\angle{IEB}$.

Hence, we have $$x=\angle{DEA}=\angle{DEI}-\angle{AEI}=\angle{DBI}-\angle{AEI}=30^\circ-10^\circ=\color{red}{20^\circ}.$$