parmenides51 wrote:
A rhombus $ABCD$ is given with $\angle BAD = 60^o$ . Point $P$ lies inside the rhombus such that $BP = 1$, $DP = 2$, $CP = 3$. Determine the length of the segment $AP$.
Join diagonal $BD$ then observe that $\triangle ADB$ is equilateral then take a point $P'$ such that $\triangle ADP\equiv\triangle DBP'$ [here $\equiv$ is a sign used by me to show congruent $\triangle $]
join $PP'$ then observe that $\triangle DPP'$ is also equilateral with side $DP=PP'=P'D=2$ then also observe that $\triangle P'DC\equiv\triangle DPB$ hence $PB=P'C=1$
Claim1 that $P', P, C$ is collinear
Proof - : it is quite simple just observe that points $P, P', C$ will either be collinear or it will form a triangle but by triangle enequality it can be proven that $\triangle P'PC$ doesn't exist also as $PP'+PC=2+1=3$ so $PP'C$ is collinear
Claim2 that $\angle DP'C=120^0$
Proof-: as $PP'C$ lie on a line (claim 1) and $\angle DP'P=60$ so $\angle DP'C=120^0$
Now just look at $\triangle PP'B$ with side length $PB=1, PP'=2, PB=AP$ and $\angle PP'C=60^0$ hence apply cosine rule in $PP'B$ and find the length of $PB=AP$ !!