Question

please help brainly warriors please

Answer 1

Refer the attachment for figure.



Given that ABCD is a ||gm.

=> AB = CD (opposite sides are equal in a ||gm)

=> 1/2AB = 1/2CD

=> BX = DY


also, since ABCD is a ||gm

=> AB || CD

=> BX || DY (Since BX and DY are parts of them)

Now in Quadrilateral BXDY,

DY = BX
DY || BX

Since opposite sides are equal and parallel,

=> BXDY is a ||gm

=> DX || BY (opposite sides are parallel in a ||gm)

Since DX || BY

=> WX || YZ (since they are a part of them)


Now in ∆WXZ and ∆WYZ,

angle ZWY = angle XZW (Since WX||YZ, alternate angles)

WZ = WZ (common)

angle WZY = angle XWZ (Alternate angles)

So ∆XWZ is congruent to ∆YZW (ASA)

=> WX = YZ (c.p.c.t)


Now WX = YZ
WX || YZ

=> XZWY is a ||gm (since opposite sides are equal and parallel)


Hence Proved :)