Division of area of a triangle

Since $[APB]:[BDP]=21:1$ , then $AP:PD=[APB]:[BDP]=21:1$ by area and base relationships. Then $[APC]:[DPC]=AP:PD=21:1$ hence $[APC]=16 \times 21=336$ .

From the fact that $BDP$ has the same altitude as $BCP$ from the vertex $P$ , and that $BDA$ has the same altitude as $BCA$ from the vertex $A$ , we have that: $BD:DC= [BDP] : [BCP]=[BDA]:[BCA]$ Therefore $[BCA]=\frac{[BCP][BDA]}{[BDP]}=\frac{26 \cdot 220}{10}=572$ Now we notice that $[APC]=[BCA]-[BDP]+[DPC]+[APB]=\fbox{336}$

The area of a triangle is proportional to the length of its base. Thus the ratio of two triangles' areas is the same as the ratio two triangles' bases. Thus $\frac{[ADB]}{[ADC]}=\frac{BD}{DC}=\frac{[BDP]}{[CDP]}=\frac{10}{16}$ . $[ADB]=[APB]+[BDP]=220$ , so $[ADC]=[ADB] \times \frac{16}{10}=352$ and $[APC]=[ADC]-[DPC]=336$ .

Notice that the distance from $B$ to $PD$ is the same as the distance from $B$ to $AD$ since $APD$ is a straight line. Thus, triangles $APB$ and $PDB$ have the same height with their base on the line $APD$ , so the ratio of their area is equal to the ratio of their base, i.e. $\frac { [APB] } {[PDB] } = \frac {AP} {PD}$ . Similarly, triangles $APC$ and $PDC$ have the same height with their base on the line $APD$ , so the ratio of their area is equal to the ratio of their base, so $\frac {[APC]} {[PDC] } = \frac {AP} {PD}$ . These two equations give us $\frac { [APB] } {[PDB] } = \frac {AP} {PD} = \frac {[APC]} {[PDC] }$ $\Rightarrow [APC] = \frac { [APB] \times [PDC] } { [PDB] } = \frac { 210 \times 16 } {10 } = 336$ .

In this special case, AD is perpendicular to CB, BE is perpendicular to CA, and CF is perpendicular to AB. Let’s say the length of DP is 2. Because of this, DC must be 16, and BD must be 10. Since [APB]+[PDB]=220, and the formula for area of a triangle is (bh)/2, the height must be 44. To find [ABC] you must use the formula and you get [ABC] which is 572. When you subtract [ABP]+[BDP]+[CDP] from [ABC], you get 336, the answer.