Question

name: brenden tw

1. given the parallelogram ebcd, eb = 16,

ed = 25, bf = 11, ec = 34,
m∠bed = 55°, m∠cdb = 67°, and
m∠bce = 24°. determine the missing measures.
bc =
m∠edc =
bd =
m∠ebd =
fc =
m∠bec =
cd =
m∠dbc =

Explanation:

Step1: Properties of parallelogram

In a parallelogram \(EBCD\), opposite sides are equal. So \(BC = ED\), \(CD=EB\).
Since \(ED = 25\), then \(BC = 25\). Since \(EB = 16\), then \(CD = 16\).

Step2: Diagonals of parallelogram

The diagonals of a parallelogram bisect each other. Given \(BF = 11\), then \(BD=2BF\). So \(BD = 2\times11=22\). Given \(EC = 34\), then \(FC=\frac{1}{2}EC\). So \(FC=\frac{34}{2}=17\).

Step3: Angles in parallelogram

Since \(EB\parallel CD\), \(\angle EBD=\angle CDB\) (alternate - interior angles). So \(m\angle EBD = 67^{\circ}\).
In \(\triangle BEC\), using the angle - sum property of a triangle. We know that in parallelogram \(EBCD\), \(EB\parallel CD\) and \(EC\) is a transversal. \(m\angle BEC=m\angle BCE\) (alternate - interior angles). So \(m\angle BEC = 24^{\circ}\).
For \(m\angle EDC\): In parallelogram \(EBCD\), \(m\angle EDC + m\angle BED+m\angle EBD=180^{\circ}\) (angle - sum property of a triangle in \(\triangle BED\) and \(EB\parallel CD\)). Substitute \(m\angle BED = 55^{\circ}\) and \(m\angle EBD = 67^{\circ}\). Then \(m\angle EDC=180-(55 + 67)=58^{\circ}\).
For \(m\angle DBC\): In parallelogram \(EBCD\), \(m\angle DBC=180-(m\angle BCD + m\angle BDC)\). First, \(m\angle BCD=m\angle BED = 55^{\circ}\) (opposite angles of parallelogram are equal). Using the angle - sum property of \(\triangle BCD\), \(m\angle DBC=180-(55 + 67)=58^{\circ}\)

Answer:

\(BC = 25\), \(BD = 22\), \(FC = 17\), \(CD = 16\), \(m\angle EDC = 58^{\circ}\), \(m\angle EBD = 67^{\circ}\), \(m\angle BEC = 24^{\circ}\), \(m\angle DBC = 58^{\circ}\)