Let's solve it by induction on $n$. The base case $n=2$ is the most difficulty part of the solution, let's see why it is a sufficient condition for the problem to be true. Suppose we know the problem to be true for $n=2,3,...,k$, then let's show how to choose $k+1$ triangles satisfying a) through c). Suppose wlog that $S(ABC) \ge \frac{1}{2}S(ABCD)$, then there is $X$ point in the segment $BC$ such that $$S(ABX)=\frac{4}{4k+5}S(ABCD) \le \frac{4}{13}S(ABCD)<\frac{1}{2}S(ABCD).$$Now we know that one can choose $k$ triangles in $AXCD$ satisfying a) through c), so each triangle will have an area at least $$\frac{4k}{4k+1}S(AXCD) = \frac{4k}{4k+1}\cdot\frac{4k+1}{4k+5}S(ABCD) = S(ABX).$$Now minimize these last $k$ triangles via a homotety of the appropriate ratio, and centered inside each triangle, so that all have the same area as $ABX$, and there is no superposition. At last, see that the total area covered by the $k+1$ triangles is in fact $\frac{4k+1}{4k+5}S(ABCD)$. Now for the base case. Suppose wlog $$S(BCD) = \min(S(ABC), S(BCD),S(CDA),S(DAB)).$$Notice that the locus of points $P$ such that $S(DAP)=S(ABP)$ contains the line passing though $A$ and $M$ midpoint of $BD$ as in fact $S(DAM)=S(ABM)$, now let $h_D$ and $h_B$ be the heights through $D$ and $B$ of the triangles $DAM$ and $ABM$, respectively, notice that $h_B=h_D$, now if $P \in AM$, $$S(DAP) = \frac{1}{2}AP\cdot h_D = \frac{1}{2}AP\cdot h_B=S(ABP).$$Let $Q$, be the point over the perimeter of $ABCD$ in which the line $AM$ passes through, that is not $A$. $Q$ is either on $BC$ or $CD$, wlog $Q$ on $CD$. Now let $$S(BCD)=x\cdot S(ABCD), S(ABC)=y\cdot S(ABCD),$$we also have $$S(DAB) = (1-x)S(ABCD), S(CDA) = (1-y)S(ABCD).$$By definition of $x$, we have $x\le(1-x) \implies x \le \frac{1}{2}$. We can prove that $y\le \frac{1}{2}$ as otherwise $1-x \ge y >\frac{1}{2}$ and we get $$S(DAB), S(ABC) > \frac{1}{2}S(ABCD) \implies S(ABQ) > \frac{1}{2}S(ABCD),$$as the area of $ABX$ varies monotonically when $X$ covers $CD$. But it is known that $$S(ABQ) = S(DAQ) \implies S(ABQD) > S(ABCD)$$which is clearly false. Looking at $ABQ$ and $DAQ$ as the $2$ triangles of choice, conditions a) and b) are met. We only need $$S(ABQD) \ge \frac{8}{9}S(ABCD) \iff S(BQC) \le \frac{1}{9}S(ABCD).$$Let $$\alpha = \frac{CQ}{CD},$$then $$S(ADQ) = \alpha \cdot S(ADC) = (1-\alpha)(1-y)S(ABCD) \implies S(ABQD) = 2(1-\alpha)(1-y)S(ABCD).$$At the same time $$S(BQC) = \alpha S(BCD) = \alpha xS(ABCD) ,$$summing those up we get $$1= 2(1-\alpha)(1-y)+\alpha x \implies \alpha = \frac{1-2y}{2-2y-x}.$$Now $$S(BQC) = \frac{x-2xy}{2-2y-x}S(ABCD),$$and it suffices that $$ \frac{x-2xy}{2-2y-x} \le \frac{1}{9} \iff 9x-18xy \le 2-2y-x \iff x \le \frac{1-y}{5-9y},$$as both $2-2x-y$ and $5-9x$ are positive, because $x\le y\le \frac{1}{2}$. That last inequality is true as $$y \le \frac{1-y}{5-9y} \iff y(5-9y)-(1-y) \le 0 \iff -9y^2+6y-1 \le 0 \iff (3y-1)^2 \ge 0,$$and by definition $x \le y$.

solved from here