Question

challenging to solve this

Answer 1

is anyone

Answer 2

@SolomonZelman @ganeshie8 @amistre64 @amistre64 @mathstudent55

Answer 3

@Hero @OrangeMaster @ankit042

Answer 4

I want to leave this here so I can see better

Answer 5

ya fig is right but ans

Answer 6

@ForgetMeNot
@JungHyunRan @Drsuz98

Answer 7

@Ready1917 @veehype @SolomonZelman @CGGURUMANJUNATH @ChasingMaggie @sleepyjess @amistre64

Answer 8

the attachment isn't working for me. And in fact even if it did :P

Answer 9

ohhhhh

Answer 10

u can use other figures

Answer 11

1. \(\Large \angle ACB = 180^\circ-(10^\circ+70^\circ)-(60^\circ+20^\circ) = 20^\circ\)
\(\Large \angle AEB = 180^\circ-60^\circ-(50^\circ+30^\circ) = 40^\circ\)

2.Draw a line from point \(\Large E\) parallel to \(\Large AB\), labeling the intersection with \(\Large AC\) as a new point \(\Large F\) and conclude \(\Large ΔCEF∼ΔABC\),
\(\Large \begin{align}
\angle CEF &= \angle CBA = 50^\circ+30^\circ = 80^\circ\\
\angle FEB &= 180^\circ-80^\circ = 100^\circ\\
\angle AEF &= 100^\circ-40^\circ = 60^\circ\\
\angle CFE &= CAB = 60^\circ+20^\circ = 80^\circ\\
\angle EFA &= 180^\circ-80^\circ = 100^\circ
\end{align}\)

3.Draw a line \(\Large FB\) labeling the intersection with \(\Large AE\) as a new point \(\Large G\) and conclude \(\Large ΔAFE≅ΔBEF\),
\(\Large \begin{align}
\angle AFB &=\angle BEA = 40^\circ\\
\angle BFE &= \angle AEF = 60^\circ\\
\angle FGE &= 180^\circ-60^\circ-60^\circ = 60^\circ = \angle AGB\\
\angle ABG &= 180^\circ-60^\circ-60^\circ = 60^\circ\\
\end{align}\)

4.Draw a line \(\Large DG\). Since \(\Large AD=AB\) (leg of isosceles) and \(\Large AG=AB\) (leg of equilateral), conclude \(\Large AD=AG\), \(\Large ΔDAG\) is isosceles and
\(\Large \angle ADG =\angle AGD = \frac{180^\circ-20^\circ}2 = 80^\circ\).

5.Since \(\Large ∠DGF=180^∘−80^∘−60^∘=40^∘\), conclude ΔFDG (with two \(\Large 40^∘\) angles) is isosceles, so \(\Large DF=DG\).

6.With \(\Large EF=EG\) (legs of equilateral) and \(\Large DE=DE\) (same line segment) conclude \(\Large ΔDEF≅ΔDEG\) by side-side-side rule, \(\Large ∠DEF=∠DEG=x\), and
\(\Large \angle FEG = 60^\circ = x+x\quad\Rightarrow\quad \large\color{blue}{x=30^\circ}.\)

Answer 12

great right answer