On the sides of triangle $ABC$, isosceles right-angled triangles $AUB, CVB$, and $AWC$ are placed. These three triangles have their right angles at vertices $U, V$ , and $W$, respectively. Triangle $AUB$ lies completely inside triangle $ABC$ and triangles $CVB$ and $AWC$ lie completely outside $ABC$. See the figure. Prove that quadrilateral $UVCW$ is a parallelogram. [asy][asy] import markers; unitsize(1.5 cm); pair A, B, C, U, V, W; A = (0,0); B = (2,0); C = (1.7,2.5); U = (B + rotate(90,A)*(B))/2; V = (B + rotate(90,C)*(B))/2; W = (C + rotate(90,A)*(C))/2; draw(A--B--C--cycle); draw(A--W, StickIntervalMarker(1,1,size=2mm)); draw(C--W, StickIntervalMarker(1,1,size=2mm)); draw(B--V, StickIntervalMarker(1,2,size=2mm)); draw(C--V, StickIntervalMarker(1,2,size=2mm)); draw(A--U, StickIntervalMarker(1,3,size=2mm)); draw(B--U, StickIntervalMarker(1,3,size=2mm)); draw(rightanglemark(A,U,B,5)); draw(rightanglemark(B,V,C,5)); draw(rightanglemark(A,W,C,5)); dot("$A$", A, S); dot("$B$", B, S); dot("$C$", C, N); dot("$U$", U, NE); dot("$V$", V, NE); dot("$W$", W, NW); [/asy][/asy]
Problem
Source: Dutch NMO 2014 p2 seniors
Tags: geometry, parallelogram, right triangle, isosceles, Isosceles Triangle
07.09.2019 16:40
Dear Mathlinkers, http://jl.ayme.pagesperso-orange.fr/Docs/Triangles%20adjacents.pdf p. 14-18. Sincerely Jean-Louis
25.01.2021 13:19
Nice Problem! The strategy used here is a unique one (I call it changing the statement, a technique which might not appear alot in your Geometry problem solving strategies!) Notice point U is a unique point in this problem, and due to this fact, we can change the problem as following: * We have triangle $ABC$, points $V$ and $W$, but we CHOOSE point $U'$ in a way that $CWU'V$ IS a parallelogram. (notice $U$ is UNIQUE due to way way it's constructed: the encountring of the 2 lines which make a $45$ $^{\circ}$angle with the side $AB$ of the triangle) Now, it's simple! Call $\angle U'VB = \alpha$ and $\angle U'VC = \beta$ so $\alpha$ $+$ $\beta$ = $\ 90$$^{\circ}$ Try to calculate $AU'B$ by calculating $ 360^{\circ}$ - ( $\angle$ $VU'B$ + $\angle$ $VU'W$ + $\angle$ $AU'W$) and notice that triangles $BU'V$ and $AU'W$ are equal. That is all you need to do $\blacksquare$