Question

In AABC, seg AD 1 Seg BC.
Prove AB square + CD square = BD squaer + AC square ​

Answer 1

Answer:

Step-by-step explanation:

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