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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 $.

Diagram

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?

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    $\begingroup$ What if $AD = BC$ in another way, like a parallelogram? $\endgroup$ – peterwhy Aug 3 '17 at 11:43
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    $\begingroup$ Proving $\angle ADC = \angle BCD$ is as difficult as proving $\angle A = \angle B$, by symmetry. $\endgroup$ – peterwhy Aug 3 '17 at 11:56
  • $\begingroup$ right.......... $\endgroup$ – Soumee Aug 3 '17 at 11:57
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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 $\angle A+\angle B=180^\circ$.

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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!

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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?).

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  • $\begingroup$ 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)? $\endgroup$ – Soumee Aug 3 '17 at 12:04
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    $\begingroup$ 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. $\endgroup$ – farruhota Aug 3 '17 at 12:19

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