Let $ABCD$ be a convex quadrilateral with $\angle{BAD} = 2\angle{BCD}$ and $AB = AD$. Let $P$ be a point such that $ABCP$ is a parallelogram. Prove that $CP = DP$.
Problem
Source: Sharygin Finals 2022 8.1
Tags: geometry, geometric transformation, reflection, circumcircle
Tsikaloudakis
04.10.2022 13:16
Ασκηση 138 \[ \begin{array}{l} Let\mathop {}\limits_{}^{} MH\mathop {}\limits_{}^{} {\rm{bicect}}\mathop {}\limits_{}^{} {\rm{ perpenticular }}\mathop {}\limits_{}^{} of\mathop {}\limits_{}^{} BC\mathop {}\limits_{}^{} and\mathop {}\limits_{}^{} BH = AB. \\ AD = AB = HC = PC\mathop {}\limits_{}^{} .\mathop {}\limits_{}^{} \\ AP//BC\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} H\hat KC = \varphi . \\ B\hat CD = \varphi \mathop {}\limits_{}^{} ,\mathop {}\limits_{}^{} D\hat AB = 2\varphi . \\ D\hat AP = \omega \mathop {}\limits_{}^{} ,\mathop {}\limits_{}^{} H\hat AB = \eta \mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} \omega + \eta = \varphi \mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} \\ H\hat BC = \omega \mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} tr.BHC = tr.ADP\mathop {}\limits_{}^{} \Rightarrow \mathop {}\limits_{}^{} PD = AD = PC \\ \end{array} \]
Attachments:
p.lazarov06
11.04.2024 15:28
Let E be such that $EBCD$ is a parallelogram. Then $A$ is the circumcircle of $EBD$. We will use complex numbers. Let $A$ be the origin. We get that $C=b+d-e$ and $P=C+A-B=d-e$. We want to prove that $\mid P-C\mid = \mid P-D\mid $ or $\mid -c\mid =\mid -e\mid $ which is clearly so.
InterLoop
21.07.2024 17:34
Construct the reflection of $A$ about $BD$, $X$ and note that it is the circumcenter of $\triangle CBD$. Thus if $Y$ was a point on the line through $C$ parallel to $BD$ and $H$ the foot from $C$ to $BD$, we would have $$\angle XCY = \angle HCY - \angle HCX = 90- (C + 2B - 180) = 270- C - 2B$$$$\angle PCY = \angle ABD = 90- C \implies \angle XCP = \angle XCY - \angle PCY = 180 - 2B$$However $\angle XCD = 90- B$, thus $CD$ bisects $\angle XCP$. We know $CX = CP$ too, so $X$ must be the reflection of $P$ in $CD$, but since $XD = CX = CP$, $DP$ is also equal to all of these values, and we are done.
Math_legendno12
04.08.2024 17:43
Construct A' such that $ABA'D$ is a parallelogram. From $<BA'D=<BAD=2<BCD$, and $A'B=AD=AB=A'D$, we have $C$ lies on circle with centre $A'$ and radius $A'B=A'D$ Implying $A'B=A'D=A'C$ Also, $APCB$ parallelogram, $ADA'B$ parallelogram, implies $DA'CP$ parallelogram (true in general) as $CP // A'D$ and $CP=AB=A'D$ Therefore $PC=A'D=A'C=P$ as required. Remark: I coin this construction of a 2nd parallelogram to gift a 3rd one for free 'completing the parallelogram'. This is also seen in constructions where you shift entire triangles from opposite sides of a parallelogram.