Question

in the rectangle below, bd = 8x - 6, ae = 21, and m∠eba = 35°. find the value of x and m∠ecb.

Explanation:

Step1: Recall rectangle diagonals properties

In a rectangle, diagonals are equal and bisect each other. So \( AC = BD \) and \( AE = \frac{1}{2}AC \), \( BE = \frac{1}{2}BD \). Since \( AC = BD \), then \( AE = \frac{1}{2}BD \). Given \( AE = 21 \), so \( BD = 2\times AE = 2\times21 = 42 \).

Step2: Solve for \( x \)

We know \( BD = 8x - 6 \) and \( BD = 42 \). So set up the equation \( 8x - 6 = 42 \). Add 6 to both sides: \( 8x = 42 + 6 = 48 \). Divide both sides by 8: \( x = \frac{48}{8} = 6 \).

Step3: Find \( m\angle ECB \)

In rectangle \( ABCD \), \( AB \parallel CD \), \( \angle ABC = 90^\circ \). Given \( m\angle EBA = 35^\circ \), so \( m\angle EBC = 90^\circ - 35^\circ = 55^\circ \). Also, in rectangle, diagonals bisect each other and \( EB = EC \) (since \( E \) is midpoint of diagonals), so triangle \( EBC \) is isosceles with \( EB = EC \). Thus \( \angle ECB = \angle EBC \)? Wait, no, wait. Wait, \( AB \parallel CD \), \( \angle BAC = \angle DCA \), but also, in triangle \( ABC \), \( \angle BAC + \angle BCA = 90^\circ \). Wait, maybe better: since \( EB = EC \), triangle \( EBC \) is isosceles, but actually, \( \angle EBA = 35^\circ \), \( AB \parallel CD \), so \( \angle BAC = \angle DCA \), but also, in rectangle, \( \angle BCD = 90^\circ \), and \( \angle ECB = 90^\circ - \angle EBA \)? Wait, no. Wait, \( \angle EBA = 35^\circ \), \( \angle ABC = 90^\circ \), so \( \angle EBC = 90 - 35 = 55^\circ \). But since \( EB = EC \) (diagonals bisect each other and are equal, so \( EB = EC \)), triangle \( EBC \) is isosceles with \( EB = EC \), so \( \angle ECB = \angle EBC \)? No, that's not right. Wait, no, \( \angle EBA = 35^\circ \), \( AB = CD \), \( AD = BC \), diagonals \( AC = BD \), \( E \) is midpoint. So \( \angle BAC = \angle EBA = 35^\circ \) (since \( EA = EB \), because diagonals bisect each other and \( AC = BD \), so \( EA = EB \)). Wait, \( EA = EB \), so triangle \( EAB \) is isosceles with \( EA = EB \), so \( \angle EAB = \angle EBA = 35^\circ \). Then in triangle \( ABC \), \( \angle BAC = 35^\circ \), \( \angle ABC = 90^\circ \), so \( \angle BCA = 90^\circ - 35^\circ = 55^\circ \). So \( m\angle ECB = 55^\circ \).

Answer:

\( x = 6 \), \( m\angle ECB = 55^\circ \)