geometry question?

Question

a quadrilaterial is drawn in such a way that three of its sides are congruent, and the 4th side is congruent to both of the diagonals. Find the mesures of the vertex angles in the quad.

its been a couple years since i've done any geometry, if you could help out, or just jog my memory, it would be great.

Answer 1

The figure we are talking about is an isosceles trapezoid. The lower base is equal to the two diagonals and the upper base and the two sides are all equal in length.

Call the trapezoid ABCD with AB being the lower base and cd bbeing the upper base. The two sides which are both equal to the upper base are BC and AD. Also notte that AB parallel to CD.

The diagonals are AC and BD.

It turns out that the two base angles are bisected by each of the diagonals.Since the base angles are equal then they are each divided into two equal angles. Call each of them angle 1. also By the parallel postulate, angle CDB and angle must also be equal to angle 1.

Now for the base to be the same length as the diagonals, triangle
ABC and Triangle must both be isosceles. This will be true only if angle ADB=.twice angle 1 and angle ACB = twice angle 1.

Therefore the two base angles add up to 4 angle 1s and the agles C and D are both = 3 angle ones. Thus the four vertex angles add up to 10 angle 1s. Since the total of all angles in a quadrilateral is 360 degrees, the angle 1 must 360/10 =36 degrees.

Thus each base angle = 2 angle 1s = 2*36 =72 degrees and each of the upper angles must be equal to 3 angle 1s =3*36=108 degrees.

In summary angle A and angle B = 72 degrees and
angles C and D both = 108 degrees.

Hope this was not too confusing. It will help if you draw a picture.

Answer 2

if the fourth side is congurent to one of the sides then they all are congruent. so itkink all of the angles would be 90 degrees

i hope this helps

Answer 3

draw a picture, it doesn't seem possible to construct