Question

hey guys help me
if u solve this question correctly i will mark brainliest answer
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Answer 1

Answer:

Step-by-step explanation:

See... Its a lengthy question

In a rhombus, all sides are equal, and A.B,C,D are midpoints, hence,

WX/2=XY/2=YZ/2=WZ/2=ZC=CY=YB=BX.....

now,

∠AWD=180°-∠DZC since oppsoite angles are suplementary

Let ∠AWD be x, so ∠DZC=180°-x

Taking in mind sum of all angles in a triangle is 180,

in triangle AWD, AW=WD, so ∠DAW=∠ADW

∠ADW=(180°-x)/2

in traingle DZC, DZ=ZC, so ∠ZDC=∠ZCD

∠ZDC={180°-(180°-x)}/2=x/2

again,

in line WZ, ∠ZDC+∠CDA+∠ADW=180°

so, ∠CDA=180°- {(180°-x)/2} -x/2 = 90°..........(i)

now,

In △ZDC and △ABX,

ZD=AX [proved]

∠DZC=∠AXB [opposite angles in a rhombus are equal]

ZC=BX [proved]

Thus, both the triangles are congruent.

So, by c.p.c.t., CD=AB........(ii)

Similarly, in △AWD and △YBC, BC=AD.........(iii)

From (i), (ii) and (iii), we get that ABCD is a quad. in which opposite sides are equal and one of the angles is 90°,

Hence, ABCD is a rectangle (proved)