Need help with Triangle Area Problems

Question

Hello, I need help with the following two questions:

1)

Construct triangle ABC such that BC is of length 1 centimetre, A=100degrees, B=60degrees, C=20degrees. Let D be the midpoint of AC and let E be the point in BC such that angle EDC is of 80 degrees.

Find the sum of the area of ABC and twice the area of CDE.
(I got up to the point where I construct a new point F such that FAC is 80 degrees and then FE=EC=CD=DA, and then CDE is 1/4 the area of FAC, what should I do now?)

2)

Triangle ABC is an acute angle triangle (where all angles are less than 90 degrees). D, E and F lies in BC, CA and AB, respectively such that AD is an altitude of a triangle, BE is a median, and CF is an angle bisector of C. Let M be a point where CF meets AD and let N be a point where CF meets BE. MN is of length 1 unit, FM is of length 2 units and NC is of length 3 units.

Find the area of triangle ABC.
(I'm drawing a blank here, all i know is that CF is 6 units long, because M is closer to F than N. Ceva's theorem doesn't help, angle bisector theorem doesn't help.)

Answer 1

1) Given that,  in ΔABC, BC = 1 centimetre, ∠A=100º, ∠B=60º, ∠C=20º and also that

AD=DC, ∠EDC=80º

Hence, in ΔEDC, ∠CED=180º-∠EDC- ∠C=180º- 80º-20º= 80º and ∠CED=∠EDC => DC=CE=(1/2)CA

We know that area of ΔABC = (1/2)CA*BC*sinC= (1/2)CA*sinC [since BC = 1 centimetre]

Also, area of ΔEDC=(1/2)DC*CE*sinC = (1/2)*(1/2)CA*(1/2)CA*sinC [since  DC=CE=(1/2)CA] 

Hence, area of ΔEDC= (1/2)*(1/2)CA*(area of ΔABC)

=> twice the area of ΔEDC= 2*(1/2)*(1/2)CA*(area of ΔABC)=(1/2)CA*((1/2)CA*sinC)={(1/2)CA}²*sinC

Therefore the sum of the area of ΔABC and twice the area of ΔCDE = (1/2)CA*sinC + {(1/2)CA}²*sinC ={1+(1/2)CA}}(1/2)CA*sinC

 

2) In ΔACF N is mid point of CF and E is mid point of CA 

Therefore, NE must be parallel to AF which implies that BE must be parallel to AB which is impossible under given conditions. 

Therefore, the problem statement needs corrections.