#7 Area Method II

The areas are in the ratio of the bases on line \(BC\). Therefore \([ABD]:[ADC]=BD:DC=4:7\).

Triangles \(ABD\) and \(ADC\) have a common altitude, so the ratio of areas equals the ratio of bases. Hence \(BD:DC=5:2\).

Triangles \(ABM\) and \(ACM\) have a common altitude from \(A\) to \(BC\). Bases \(BM\) and \(CM\) are equal, so the areas are equal.

Triangles \(ABP\) and \(ABQ\) have common base \(AB\). Their areas are in the ratio of the altitudes to this base, hence \([ABP]:[ABQ]=3:1\).

Triangles \(ABP\) and \(ACP\) have common base \(AP\). Their altitudes from \(B\) and \(C\) to \(AP\) are in the ratio \(BD:DC\). Therefore \(\frac{[ABP]}{[ACP]}=\frac{BD}{DC}\).

By the formula, \(\frac{BD}{DC}=\frac{[ABP]}{[ACP]}=\frac{6}{5}\). Hence \(BD:DC=6:5\).

We have \(\frac{BD}{DC}=\frac{[PAB]}{[PCA]}=\frac{12}{6}=2:1\). Next, \(\frac{CE}{EA}=\frac{[PBC]}{[PAB]}=\frac{9}{12}=3:4\). Finally, \(\frac{AF}{FB}=\frac{[PCA]}{[PBC]}=\frac{6}{9}=2:3\).

Triangles \(PBC\) and \(ABC\) have common base \(BC\). Hence \(\frac{[PBC]}{[ABC]}=\frac{PD}{AD}=\frac{3}{8}\). Then \(AP=AD-PD\), so \(AP:PD=5:3\).

By the formula, \(\frac{[PAB]}{[PAC]}=\frac{BM}{MC}\). Since \(M\) is the midpoint of \(BC\), \(BM=MC\). Therefore \([PAB]=[PAC]\).

We have \(\frac{BD}{DC}=\frac{[PAB]}{[PAC]}=1\). Therefore \(BD=DC\), so \(D\) is the midpoint of \(BC\).

Let \(S_A=[PBC]\), \(S_B=[PCA]\), \(S_C=[PAB]\). Then \(\frac{BD}{DC}=\frac{S_C}{S_B}\), \(\frac{CE}{EA}=\frac{S_A}{S_C}\), \(\frac{AF}{FB}=\frac{S_B}{S_A}\). Multiplying cancels everything and gives \(1\).

From \(BD:DC=2:3\), we get \(x:z=2:3\). From \(CE:EA=3:4\), we get \(y:x=3:4\). Take \(x=8\). Then \(z=12\), \(y=6\). Hence \([PAB]:[PBC]:[PCA]=8:6:12=4:3:6\).

Since \(AP:PD=4:3\), we have \(PD:AD=3:7\). Triangles \(PBC\) and \(ABC\) have common base \(BC\), so \([PBC]:[ABC]=PD:AD=3:7\).

Let \([PAB]=x\), \([PBC]=y\), \([PCA]=z\). Then \(x:z=3:5\), and \(y:x=2:3\). Take \(x=9\). Then \(z=15\), \(y=6\). Answer: \([PAB]:[PBC]:[PCA]=9:6:15=3:2:5\).

Since the cevians pass through one point, by Ceva \(\frac{5}{4}\cdot\frac{3}{5}\cdot\frac{AF}{FB}=1\). The first factors give \(\frac{3}{4}\), so \(\frac{AF}{FB}=\frac{4}{3}\). Answer: \(AF:FB=4:3\).

Let \([PAB]=x\), \([PBC]=y\), \([PCA]=z\). From \(BD:DC=2:3\), \(x:z=2:3\). From \(CE:EA=3:2\), \(y:x=3:2\). Take \(x=2\); then \(z=3\), \(y=3\). The total area is \(8\) units, and \([PBC]=3\). Since \(P\in AD\), \(\frac{PD}{AD}=\frac{[PBC]}{[ABC]}=\frac{3}{8}\). Hence \(AP:PD=5:3\).

Let \([PAB]=x\), \([PBC]=y\), \([PCA]=z\). From \(BD:DC=3:4\), \(x:z=3:4\). From \(CE:EA=2:5\), \(y:x=2:5\). Take \(x=15\); then \(z=20\), \(y=6\), total \(41\). Since \(P\in BE\), \(\frac{PE}{BE}=\frac{[PCA]}{[ABC]}=\frac{20}{41}\). Thus \(BP:PE=21:20\).

The product is \(\frac{z}{y}\cdot\frac{x}{z}\cdot\frac{y}{x}=1\). By Ceva's theorem, lines \(AD\), \(BE\), \(CF\) are concurrent. The numbers \(x,y,z\) may be interpreted as the areas of the three small triangles around the intersection point.

Since \(PQ\parallel BC\), triangles \(APQ\) and \(ABC\) are similar with ratio \(\frac{AP}{AC}\). Areas of similar triangles are in the ratio of the squares of similarity ratios, hence \(\frac{[APQ]}{[ABC]}=\left(\frac{AP}{AC}\right)^2\).

From similarity \(APQ\sim ABC\), \(\left(\frac{AP}{AC}\right)^2=\frac{9}{25}\), hence \(\frac{AP}{AC}=\frac{3}{5}\). Then \(PC=\frac{2}{5}AC\), so \(AP:PC=3:2\).

Let \([PAB]=x\), \([PBC]=y\), \([PCA]=z\). From \(BD:DC=2:5\), \(x:z=2:5\). From \(CE:EA=3:4\), \(y:x=3:4\). Take \(x=8\); then \(z=20\), \(y=6\), total \(34\). On cevian \(AD\): \(\frac{PD}{AD}=\frac{[PBC]}{[ABC]}=\frac{6}{34}=\frac{3}{17}\), so \(AP:PD=14:3\). On cevian \(BE\): \(\frac{PE}{BE}=\frac{[PCA]}{[ABC]}=\frac{20}{34}=\frac{10}{17}\), so \(BP:PE=7:10\).

The product \(\frac{3}{4}\cdot\frac{2}{3}\cdot 2=1\), so by Ceva the cevians are concurrent. Let \(x=[PAB]\), \(y=[PBC]\), \(z=[PCA]\). From \(BD:DC=3:4\), \(x:z=3:4\). From \(CE:EA=2:3\), \(y:x=2:3\). Take \(x=9\); then \(z=12\), \(y=6\). Answer: \([PAB]:[PBC]:[PCA]=9:6:12=3:2:4\).

We have \(\frac{AP}{AC}=\frac{1}{3}\), \(\frac{AQ}{AC}=\frac{2}{3}\). Therefore \([APP_1]:[ABC]=\frac{1}{9}\), while \([AQQ_1]:[ABC]=\frac{4}{9}\). Hence \([APP_1]:[AQQ_1]=1:4\).

By the formulas, \(\frac{BD}{DC}=\frac{S_C}{S_B}=\frac{15}{10}=3:2\). Next, \(\frac{CE}{EA}=\frac{S_A}{S_C}=\frac{6}{15}=2:5\). Finally, \(\frac{AF}{FB}=\frac{S_B}{S_A}=\frac{10}{6}=5:3\). Ceva check: \(\frac{3}{2}\cdot\frac{2}{5}\cdot\frac{5}{3}=1\), so the ratios are consistent with concurrence of the cevians.