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

Question

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),

$DC=DC$.

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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?

Answer 1

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 $\angle A+\angle B=180^\circ$.

Answer 2

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!

Answer 3

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?).