If ABCD is a quadrilateral in which AB || CD and AD=BC, prove that $\angle$A=$\angle$ B.

Q. Let ABCD be a quadrilateral in which AB || CD and AD=BC. Prove that $\angle$A=$\angle$ B.

My attempt:

Connecting BD and AC and trying to prove $\Delta ADC \cong \Delta BCD$.

In $\Delta ADC \text{ and }\Delta BCD$:

$AD=BC$(given),

$DC=DC$.

I need to prove $\angle ADC = \angle BCD$ in order to prove the two triangles are congruent by SAS congruency criterion. But I don't know how $\angle ADC = \angle BCD$ in the above problem?

• What if $AD = BC$ in another way, like a parallelogram? – peterwhy Aug 3 '17 at 11:43
• Proving $\angle ADC = \angle BCD$ is as difficult as proving $\angle A = \angle B$, by symmetry. – peterwhy Aug 3 '17 at 11:56
• right.......... – Soumee Aug 3 '17 at 11:57

Drop an altitude onto $AB$ from $D$ and from $C$. Since $AB\parallel DC$, the altitudes have the same length. Also $AD=BC$, so

$$\sin \angle A=\sin \angle B$$

Either $\angle A=\angle B$, or $\angle A+\angle B=180^\circ$.

Let $E\in{DC}$ such that $AE||BC$. Thus, $AECB$ is parallelogram,

which gives $AE=BC$ and from here $AE=AD$.

Thus, $\measuredangle ADE=\measuredangle AED=\measuredangle BCD$.

Now, $$\measuredangle DAB=180^{\circ}-\measuredangle ADC=180^{\circ}-\measuredangle BCD=\measuredangle ABC$$ and we are done!

Hints:

$1)$ drop altitudes $AE$ and $BF$ onto $BC$ and prove $\Delta ADE=\Delta BCF$.

$2)$ in two right angle triangles, if two corresponding sides are equal, the third sides are also equal (why?).

• Right... By RHS congruency criterion $\Delta ADE=\Delta BCF$ . But is there any property which says perpendicular distance between two parallel lines are equal?...(It can be deduced intuitively though)......So you are equating the right angle,hypotenuse and THE SIDE(THE ALTITUDE)? – Soumee Aug 3 '17 at 12:04
• for distance between parallel lines check here: en.wikipedia.org/wiki/Distance_between_two_straight_lines. for equality, when hypotenuses and one legs of right angle triangles are equal, then due to Pythagoras theorem, the third sides are uniquely defined and also equal. – farruhota Aug 3 '17 at 12:19