+0

0
54
6

been working on this for an hour and have no clue! Honestly would very much appreciate a solution. Thank you so much!!

Mar 1, 2019

#1
+223
0

Still fully working on this. The area of ABC = $$\sqrt{79251200}$$. Try to do something with the angle bisector theorem. Also, if you extend CI to AB at Z, AZ = 2/5 AB

Mar 1, 2019
edited by LagTho  Mar 1, 2019
edited by LagTho  Mar 1, 2019
#2
+99329
+1

I know that these geometry questions can be infuriating and I hope someone gives you a helpful hint.     But NOT a solution.

(Although sometimes that is very difficult for answerers to do - I mean it can be difficult to give hints without giving solutions)

Mar 1, 2019
edited by Melody  Mar 1, 2019
#4
+99329
+1

I think this question is MUCH easier than it first appears to be.

The little tirangle and the big one are similar triangles BUT you do not need to use this fact.

1) Mark everything that you know on the diagram.

2) let distance  MI=x and IN=y    that will make it easier

3) Find missing angles using facts about angles on a transversal through  parallel lines.

4) use these angles to deduce facts about side lengths.

5) You should now have algebraic expressions for lengths AM and AN and MN

6) add those lengths together and see what you get :))

Please no one answer over the top of me, I am trying to teach.

If you get stuck you can ask specific questions.   :)

Mar 2, 2019
edited by Melody  Mar 2, 2019