Question

wxyz is a quadilatera whose digonal intersect each other at point o such that ow=ox=oz if angle owx=50 degree then find the measure of angle ozw

Answer 1

The sol. is ====> In ΔOWX, OW=OX,................. ∴ ∠OWX = ∠OXW = ∠50° ( angles opp. to equal sides are equal ).................... ∴ ∠O+∠W+ ∠X = 180 ( angle sum property of a Δ).................. ∴ 50+50+∠O = 180.................... 100+∠O = 180.................. ∠O = 80°...................... Now in ΔOWZ, ∠ZOW + ∠WOX = 180 ( linear pair )................ ∠ZOW + 80 = 180.................. ∠ZOW = 100................. In ΔOZW, ∵ OZ = OW,................. Let ∠Z= x,................. ∴∠W= x ( angles opp. to equal sides are equal ) ,................. Now, ∠O+∠Z+∠W = 180( angle sum property of a Δ).................... 100+2x = 180,................... 2x = 80,................. x = 80/2,................. ∴ x = 40°.................. ∴ OZW = 40°................

Answer 2

see diagram.

    Given OW = OX,  so In ΔOXW, ∠OWX = ∠OXW = 50°
              OW = OZ ,   so in ΔOWZ, ∠OWZ = ∠OZW
 
In ΔWXZ ,  Sum of angles:
    50° + (50° + ∠OWZ)+ ∠OZW = 180°
   => ∠OWZ = ∠OZW = 80/2 = 40°

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Cyclic quadrilateral method:

     As OX= OW = OZ, they are on a circle with OW = OZ= OX as the radius. Obviously XZ = diameter. As diameter makes 90° at the circumference,  ∠OWZ = 90-50° = 40°.

   Since  OW = OZ, OWZ is an isosceles triangle  and ∠OZW = 40°