We are given a triangle $ABC$ such that $\angle BAC < 90^{\circ}$. The point $D$ is on the opposite side of the line $AB$ to $C$ such that $|AD| = |BD|$ and $\angle ADB = 90^{\circ}$. Similarly, the point $E$ is on the opposite side of $AC$ to $B$ such that $|AE| = |CE|$ and $\angle AEC = 90^{\circ}$. The point $X$ is such that $ADXE$ is a parallelogram. Prove that $|BX| = |CX|$.
Problem
Source: Irish Mathematical Olympiad 2023 Problem 1
Tags: geometry, parallelogram, congruent triangles
Mathlover_1
15.05.2023 11:57
$|EX|$$=$$|DB|$, $|EC|=|XD|$ and $\angle XEC = \angle XDB$ then we get $\triangle XEC$ and $\triangle BDX$ congruent from there we get $|CX|=|XB|$
lifeismathematics
24.05.2023 12:54
$\textbf{\textcolor{red}{Claim:-}} \triangle{BDX} \cong \triangle{XEC}$ $\textbf{\textcolor{green}{Proof:-}}$ we notice that $|AE|=|DX|$ and also, $|DB|=|AD|=|XE|$ and also $\angle{DAB}=45^{\circ}=\angle{EAC}$ so we have $\angle{BDX}=\angle{XEC}=\angle{A}$ and hence $\triangle{BDX} \cong \triangle{XEC}$ hence claim follows $\square$ so from claim we have $|BX|=|CX|$ $\blacksquare$
Attachments:
sami1618
11.05.2024 02:14
Construct squares $ABFG$ and $ACHI$ outside the triangle. It is well known that that, $X'$, the midpoint of $FH$, is equidistant from $B$ and $C$. However since $D$ and $E$ are the midpoints of $AF$ and $AH$, respectively, $ADX'E$ is a parallelogram so we must have $X=X'$ concluding the proof.
Attachments:
