BdMO 2015 National Secondary Problem 6

Question

A Trapezoid $ABCD$ has sides $AB = 92, BC = 50, CD = 19\; and\; AD = 70$ with $AB \mid \mid CD$. Let $P$ be a point on $AB$ such that the lengths of the perpendicular lines from $P$ to $AD$ and $BC$ are same. Find the length of $AP$.

My Work

I tries expressing $AP_1\; and\; AP_2$ in terms of $AP$ but failed. It is leading to too much complexity. How do I start?
Source: BdMO 2015 National Secondary Problem 6

Answer 1

let $$PP_1=PP_2=a$$ and the angles in $A$ $\alpha$ and in $B$ $\beta$ then we have $$\sin(\alpha)=\frac{a}{AP}$$ and $$\sin(\beta)=\frac{a}{92-AP}$$ further let $h$ the hight of the trapez and from this we get $$\frac{a}{92-AP}=\frac{h}{50}$$ and $$\frac{a}{AP}=\frac{h}{70}$$ from here we get $$\frac{70}{AP}=\frac{50}{92-AP}$$ from this equation we get $AP$

Answer 2

Use law of cosines to find the angle $\alpha=CEB$ and $\beta=CBA$. then call $PP2=a$, so $AB=a\cdot(\sec\alpha+\sec\beta)$ so you can find $a$ and then find $AP$

Answer 3

If you subtract the area of the triangle PCD from the trapezoid, you get $ah/2$, where $a = AB$. Let $k$ denote the length of the heights of the triangles PAD and PBC, which are assumed to be equal. Then $ah/2 = 70 k + 50 k = 120 k$. On the other hand, the triangle PAD has area $AP \cdot h/2$. Solving for $AP$ then gives its length as $92 \cdot 7/12$.