Question

in square abcd, diagonals \\(\overline{ac}\\) and \\(\overline{db}\\) intersect at point e, \\(m\angle ebc = (3x + 9y)\degree\\), and \\(m\angle eab = (5x + 5y)\degree\\).

what are the values of x and y?

\\(x = \\)

\\(y = \\)

Explanation:

Step1: Recall properties of square diagonals

In a square, diagonals bisect the angles and are equal in length, and they intersect at 90°. Also, the diagonal of a square bisects the right angles of the square, so \( \angle EAB = 45^\circ \) and \( \angle EBC = 45^\circ \) (since in square \( ABCD \), \( \angle ABC = 90^\circ \), and diagonal \( BD \) bisects it, so \( \angle EBC=\frac{90^\circ}{2} = 45^\circ \); similarly, diagonal \( AC \) bisects \( \angle DAB = 90^\circ \), so \( \angle EAB = 45^\circ \)).

So we have two equations:

1. \( 3x + 9y=45 \) (since \( m\angle EBC = 45^\circ \))
2. \( 5x + 5y=45 \) (since \( m\angle EAB = 45^\circ \))

Step2: Simplify the second equation

Divide the second equation \( 5x + 5y = 45 \) by 5:
\( x + y=9 \)
So we can express \( x = 9 - y \)

Step3: Substitute \( x = 9 - y \) into the first equation

Substitute \( x = 9 - y \) into \( 3x + 9y = 45 \):
\( 3(9 - y)+9y = 45 \)
Expand the left - hand side: \( 27-3y + 9y=45 \)
Combine like terms: \( 27 + 6y=45 \)
Subtract 27 from both sides: \( 6y=45 - 27=18 \)
Divide both sides by 6: \( y=\frac{18}{6}=3 \)

Step4: Find the value of \( x \)

Substitute \( y = 3 \) into \( x = 9 - y \):
\( x=9 - 3 = 6 \)

Answer:

\( x = 6 \)
\( y = 3 \)