Question

Chord AC¯¯¯¯¯¯¯¯ intersects chord DB¯¯¯¯¯¯¯¯ at point E. If the mAD=67° and mBC=87°, then find the m∠AED.(1 point)m∠AEB=°

Answer 1

To find \( m\angle AED \), we can use the property of intersecting chords. When two chords intersect, the angles that are formed are related to the arcs they subtend.

Given:
- \( m\angle ADB = 67^\circ \)
- \( m\angle BCA = 87^\circ \)

We need to find \( m\angle AEB \).

Using the property of angles formed by two intersecting chords, we have:

\[
m\angle AEB = \frac{1}{2} (m\angle ADB + m\angle BCA)
\]

Substituting the values we have:

\[
m\angle AEB = \frac{1}{2} (67^\circ + 87^\circ)
\]

Calculating the sum:

\[
67^\circ + 87^\circ = 154^\circ
\]

Now,

\[
m\angle AEB = \frac{1}{2} (154^\circ) = 77^\circ
\]

Thus, the measure of \( m\angle AEB \) is \( 77^\circ \).

Now, to find \( m\angle AED \), we also need to use the fact that \( m\angle AED \) and \( m\angle AEB \) are supplementary angles because they are linear pairs formed by the intersecting chords.

\[
m\angle AED + m\angle AEB = 180^\circ
\]

Substituting \( m\angle AEB \):

\[
m\angle AED + 77^\circ = 180^\circ
\]

Solving for \( m\angle AED \):

\[
m\angle AED = 180^\circ - 77^\circ
\]
\[
m\angle AED = 103^\circ
\]

So, the final answer is:

\[
m\angle AED = 103^\circ
\]