# How can both angles be equal?

In the given figure $ABCD$ and $EFCD$ are parallelograms on the same base $DC$. My textbook says that $\widehat{ADE}= \widehat{FCB}$. But I failed to understand why(the reason) these two angles are equal? The reason which my textbook presents is DE is parallel to CF and AD is parallel to BC.Hence,angle between DE and AD=angle between CF and BC How can both the angles be equal by the given reason in my textbook?

• If we consider the orientation, the correct equality would be angle(EDA)=angle(FCB). If the triangle $ADE$ is moved to the right with distance $DC$ , it lands exactly on the triangle $BCF$. So, the angles must coincide. Commented Nov 16, 2016 at 13:53
• This implicitly uses what the picture also suggest; that $A$, $B$, $E$ and $F$ are on a single line. Commented Nov 16, 2016 at 13:56

We need to subtract two angles:

From the definition of transversal we know that:

$\bullet \hspace{0.5 cm} \angle DAE \cong \angle CBF$

$\bullet \hspace{0.5 cm} \angle AED \cong \angle BFC$

Consider triangles $\Delta ADE, \Delta BCF$ and the fact that the sum of angles in a triangle is $180$:

$m\angle DAE+m\angle AED + m\angle ADE = 180$ and $m\angle CBF+m\angle BFC + m\angle FCB = 180$

Thus:

$m\angle DAE+m\angle AED + m\angle ADE = m\angle CBF+m\angle BFC + m\angle FCB$

Substituting values we obtain:

$m\angle CBF+m\angle BFC + m\angle ADE = m\angle CBF+m\angle BFC + m\angle FCB$

Canceling out we are left with:

$m\angle ADE = m\angle FCB$