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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
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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

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    I would start trying to "complete" the trapeze into a triangle, with the "tip" E, so that EDC is similar to EAB... – 2017-01-23
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    i think this doesn't help – 2017-01-23
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    it doesn't help. – 2017-01-23
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    where can we find these problems? – 2017-01-23
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    did you mean BdMO Problems? – 2017-01-23

3 Answers 3

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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 $AP$

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    We dont need CD = 19? – 2017-01-23
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    if my calculations are right then we don't need this $CD$ have you got the result? – 2017-01-23
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    yep... in the main question it was said that the answer is in the form $m/n$. I got this also ^_^ – 2017-01-23
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    I think, some other solution might need the length of $CD$. – 2017-01-23
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    then show us this solution – 2017-01-23
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    what? didn't get you -_- – 2017-01-23
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    i have got a solution without using $CD$ as you see – 2017-01-23
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    Yes.. You solution i.e. without using $CD$ gives same result if we use some other solution using $CD$ .. Like in @franzlemmermeyer 's solution... Since $CD||AB$ then the solution is unique. – 2017-01-23
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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 $AP$

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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 $92 \cdot 7/12$.