Question

abcd is a square. A line segment AM intersects the side BC at M and the diagonal BD at Oxford such that angle AOB =110degree.find the value of x

Answer 1

I think your question is wrong or made intentionally tricky.Anyways, taking the question as such, we shall try and solve it.

ABCD is a square, which means AB=BC=CD=DA and AC=BD.

Now E is a point on the extension of side BC and it is given AE=BD. Now the only way this is possible, is if the point E lies at C i.e. E and C are the same point, one of the vertex of the square.

Also AE (i.e. AC) intersects DC at G. But AC intersects DC at C. Hence G and C are the same points.

Therefore, C, E, G are the same points.

Thus the intersection of AE and BD is the same as the intersection of AC and BD which is the point of intersection of the diagonals of the square, given as F.

Thus FG is a finite distant from intersection of the diagonals to the vertex while GE is zero, since E and G are the same points.

Therefore FG and GE are not equal.

Answer 2

Answer : Given :In the figure ABCD is a square. A line segment DX cuts the side BC at X and the diagonal AC at O such that COD = 105 degree and OXC = y

To find :y 

 

Since DX is a line segment, then 

angle COD + angle COX = 180 degrees {Angle sum property}

=> angle COX = 180 -105 = 75 degrees........(1)

 

Since ABCD is a square and AC is a diagonal , therefore all the angles in a square is 90 degree each and AC bisects the angle in half. {property}

=> angle ACB {  also OCX } = 90 /2 =45 degrees...............(2)

 

 

In traingle COX 

=> angle OCX + angle COX  + y = 180 degree { sum of angles in a triangle is 180 degree}

=> y = 180 - (45+75) {using eq 1 and 2}

=> y = 60 degrees Answer