<u>Let's consider the facts at hand</u>:
- By Vertical Angle Theorem ⇒ ∠BCE = ∠DCF
- ∠BEC = ∠DFC
- Sides BE = DF
<u>Based on the diagram, triangles BCE and triangles DCF are similar</u>
⇒ based on the Angle-Angle theorem
⇒ since ∠BCE = ∠DCF and ∠BEC = ∠DFC
⇒ the two triangles are similar
Hope that helps!
<em>Definitions of Theorem I used:</em>
- <u><em>Vertical Angle Theorem: </em></u><em>opposite angles of two intersecting lines must be equal</em>
- <u><em>Angle-Angle Theorem:</em></u><em> if two angles of both triangles are equal, then the given triangles must be similar</em>
<em />
Answer:
![\quad x=\frac{3}{2}](https://tex.z-dn.net/?f=%5Cquad%20x%3D%5Cfrac%7B3%7D%7B2%7D)
Step-by-step explanation:
,
,
, Subtract 23 and 3x from both sides and simplify:
, Divide both sides by 2 and simplify: ![x=\frac{3}{2}](https://tex.z-dn.net/?f=x%3D%5Cfrac%7B3%7D%7B2%7D)
<em>Hope this helps!!!</em>
Answer:
2 mm
Step-by-step explanation:
Let x be the sides of the original square,
then the area of the original square is x^2
The sides of the new square is x + 6
The area is (x + 6)^2
Then (x + 6)^2 = 16*x^2
Take square root on both sides and get
x + 6 = 4x,
3x = 6
so x = 2 mm