Let $ABC$ be a triangle where a point $K$ on $AB$ such that $AK:KB=2:3$ and on the side $AC$ is a point $L$ such that $AL:LC=5:3$. The intersection of $CK$ and $BL$ is $P$. $PH \perp AB$ and $PH = 1,5$. My question is what is the length of $AB$?
$\\$ I've used Menelaus' theorem for triangles $ABL$ and $ACK$. My results are $BP:PL = 4$, $CP = PK$. I have a strong suspicion that the answer is $AB = \frac{5}{4}PH = 1.875$. I believe that we need to draw some other lines. Any hints or ideas?
I suspect that your question cannot be solved. You are right that $CP = PK$ by Menelaus theorem. We use coordinate geometry. If $A = (0,0), B = (5a,0), C = (8b,8c)$, then $L = (5b,5c)$ and $K (2a,0)$. The fact that $PH = 1.5$ simply tells us that $4c = 1.5$ hence $c = 0.375$. The length of $AB$, which is $5a$, is not determined from $c$, in other words, we can set any $a$ and $b$ and get different answer while still conforming the question. We might need additional info to crack this question.