A line parallel to the altitude of a triangle

Question

The altitude $CH$ of a triangle $ABC$ divides $AB$ into two parts $-$ $AH=14$ $cm$ and $BH=36$ $cm$. A line $a$ parallel to $CH$ divides the area of the triangle in half and intersects $AB$ at $P$. Find $AP$ and $BP$. Since $a$ is parallel to $CH$ and $CH\perp AB\Rightarrow a\perp AB.$ From here $\triangle CHB \sim \triangle NPB$ and $$\dfrac{CN}{NP}=\dfrac{BH}{BP}=\dfrac{BC}{BN}.$$ This doesn't help much. I don't see how we can use the fact that the line divides the area of triangle $ABC$ in half. Thank you in advance! Any help would be appreciated.

Answer 1

On the contrary, the similar triangles you found is the key.

Let $BP = x$ and $CH = h$. By the similar triangles, we have $NP = \dfrac {hx}{36}$.

The area of $\triangle BPN$ is $\dfrac {hx^2}{2\times 36}$, and the area of $\triangle ABC$ is $25h$, which is twice that of $\triangle BPN$.

Hence $25h = \dfrac {hx^2}{36}$, which gives $x=30$.

Answer 2

Assume the point $P$ is some proportion, $r$ of along the path of $HB$. That is to say $HP = r\cdot HB$.

Then the area of $\triangle PNB$ is $r^2\times$ the area of $\triangle HCB=r^2\times \frac 12 HB\times HC=r^2\times 18 HC$

So we have the area of $\triangle PNB = \frac 12\times$ the area $\triangle ABC = \frac 12 \times \frac 12 AB\times HC = \frac 14\times (14+36)HC = 12.5HC$.

SO we have $r^2 \times 18 HC = 12.5 HC$ and so E$r^2 = \frac {12.5}{18}= \frac {25}{36}$ and $r = \frac 56$.

So $HP = \frac 56 36=30$ and $BP = 36 -30 = 6$. And $AP = HP +AH = 14 + 30 = 44$.