Let $ABCD$ be a convex quadrilateral. Construct $S$ and $T$ on the side $AD$ and $AB$ respectively such that $AS=AT$. Construct $U$ and $V$ on the side $BC$ and $CD$ respectively such that $CU=CV$. Assume that $BT=BU$ and $ST, UV, BD$ are concurrent, prove that $AB+CD=BC+AD$.
Problem
Source: Thailand MO 2024 Day 1 P1
Tags: geometry
beansenthusiast505
10.05.2024 14:47
this feels like a proof version of an AIME qn
do menelaus on triangles $ABD$ and $BCD$ and you get $DS=DV$
sami1618
10.05.2024 17:14
Let the three lines intersect at $P$. Applying Menelaus' Theorem twice we get $$\frac{AT}{BT}\cdot\frac{BP}{DP}\cdot\frac{DS}{AS}=1$$$$\frac{BU}{CU}\cdot\frac{DP}{BP}\cdot\frac{CV}{DV}=1$$Multiplying these two statements together and using the given equalities gives $DS=DV$ finishing the problem.
rdfzlls
02.08.2024 13:08
quite easy, just needs to use menelaus