# Given a specific trapezoid, prove it is a rectangle

In quadrilateral ABCD, AB is parallel to CD. AC and BD meet at E. Points M and N are the midpoints of AE and DE, respectively. BM and BE trisect $\angle ABC$, and CE and CN trisect$\angle BCD$. Prove that ABCD is a rectangle.

I noted that $\angle ABE$ and $\angle EDC$ are congruent and constructed DP, the angle bisector of $\angle EDC$. I also thought about using angle addition to show the angles. I think angle addition and the angle sum of a triangle is the way to go, but I don't know how to implement it. It might be something along the lines of 30-60-90 triangles.

How can I continue? Please provide some hints to the end (as I struggled with this question for over two hours already). Thanks!

Hint 1: triangles ABM and CBM must be similar. (And the equivalent triangles on the other side).

Hint 2:

Angle AEB = angle CED

Hint 3:

So angle ECN = angle EBM

You may refer to the following diagram:

Consider $ΔABE$ and $ΔEDC$.

$ΔABE \sim ΔEDC$ as they have equal corresponding angles.

As you propose, construct a line from $D$ meeting $EC$ at $P$, parallel to $BM$.

$DP$, as $BM$ does, bisects $∠EDC$ and cuts $EC$ in half.

Then $NC$ and $DP$ bisect $∠DCE$ and $∠EDC$, and halve $ED$ and $EC$, respectively.

Hence $ΔEDC$ is $60$-$60$-$60$.

Why? Construct a line from $E$ meeting $DC$ at $Q$ through the intersection of $NC$ and $DP$. By sine formula,

$$\frac{\sin∠EQC}{EC}=\frac{\sin∠CEQ}{QC} \text{ and } \frac{\sin∠DQE}{ED}=\frac{\sin∠QED}{DQ}$$

$$\frac{\sin∠CEQ}{QC}=\frac{\sin∠QED}{DQ}\Longrightarrow \frac{\sin∠EQC}{EC}=\frac{\sin∠DQE}{ED}\Longrightarrow EC=ED.$$

Using the same argument, we can infer that $ΔEDC$ has equal sides.

Could you prove that $∠ADC=90^\circ$?

One interesting thing is that the intersecting point of $NC$ and $DP$ is both the incenter and the centroid of $ΔEDC$. Dr. Floor, from Ask Dr. Math, proved that a triangle being equilateral, and the centroid being incenter, is biconditional (Equilateral Triangle - Centroid/Incenter).

$\underline{\text{ Another Method:}}$

Please refer to the following figure:

Consider $ΔAED$.

By the midpoint theorem, $MN \parallel AD$ and $MN=\frac{1}{2}AD$.

Now extend $BM$ and $CN$ to meet one another at $F$, and consider $ΔBFC$, $ΔABE$ and $ΔEDC$.

$$∠FBC=∠ABE=∠EDC \text{ and } ∠FCB=∠EAB=∠ECD$$ $$\Longrightarrow ΔBFC \sim ΔABE \sim ΔEDC.$$

As $CN$ bisects $\angle ECD$ and halves $ED$, $MC$ bisects $\angle FCB$ and halves $FB$; as $BM$ bisects $\angle ABE$ and halves $AE$, $NB$ bisects $\angle FBC$ and halves $FC$.

That means $M$ and $N$ are the midpoints of $FB$ and $FC$, respectively. Hence, by the midpoint theorem, $MN\parallel BC$ and $MN=\frac{1}{2}BC$.

Thus, $AD\parallel BC$ and $AD=BC$.

Then quadrilateral $ABCD$ is a parallelogram.

Consider $ΔABM$ and $ΔCBN$; $ΔCDN$ and $ΔCBM$.

$$∠ABM=∠CBN \text{ and } ∠NCD=∠MCB$$ $$∠MAB=∠NCB \text{ and } ∠CDN=∠CBM$$ $$\Longrightarrow ΔABM \sim ΔCBN \text{ and } ΔCDN \sim ΔCBM.$$

$$\frac{AB}{DC}=1 \implies \frac{AB}{BC}=\frac{DC}{BC}.$$ That means the ratio of the corresponding sides of $ΔABM$ and $ΔCBN$ is the ratio of the corresponding sides of $ΔCDN$ and $ΔCBM$. Hence

$$\frac{AM}{NC}=\frac{NC}{MC} \implies 3AM^2=NC^2$$ $$\implies NC=\sqrt{3}AM$$

Consider $ΔEDC$.

Using the angle bisector theorem, as Rex proposed,
$$\frac{CE}{CD}=\frac{EN}{ND}=1 \implies CE=CD \implies CN \perp ED.$$

Consider $ΔENC$.

Since $EC=2AM$ and $NC=\sqrt{3}AM$ and $CN \perp ED$, $EN=\sqrt{4-3}AM=AM.$

Then $ΔENC$ is $30$-$60$-$90$. Thus, $ΔAEB$ and $ΔEDC$ are $60$-$60$-$60$. Since $AD\parallel BC$, quadrilateral $ABCD$ is a rectangle.

First, we can determine that the trisected angles are all congruent. Because of this, $\triangle$${ECD}'s median CN is an angle bisector of \triangle$${ECD}$. The angle bisector theorem tells us that $\frac{EC}{CD} = \frac{EN}{ND} = 1$, therefore $\triangle{ECD}$ is isosceles. Because of this, we can find that AB = BE and BM is perpendicular to AE. $\angle{BME} = \angle{CNE}$ and $\angle{MEB} = \angle{CEN}$, $\triangle{BEM} \sim \triangle{CEN}$. Therefore, $\angle{ECN} = \angle{EBM}$, and $\angle{ABC} = 3\angle{MBE} = 3\angle{ECN} = \angle{BCD}$. $\angle{ABC} + \angle{BCD} = 180$ degrees. Therefore, each of those two angles are right angles. Since $\angle{EBC}=\angle{ECB}$, $BE = CE$. The isosceles triangles show that $CD = CE = BE = BA$. ABCD is a parallelogram since there's a pair of parallel and congruent opposite sides. Since $\angle{ABC}$ is right, $ABCD$ is a rectangle.

• This is a complete proof, and I really like the use of angle bisector theorem, which was unbeknownst to me. May 23 '15 at 7:03
• What happened to the formatting??
– Rex
May 28 '15 at 16:14
• I don't quite know. I had rewritten \triangle{BEM} ~ \triangle{CEN} as \triangle{BEM} \sim \triangle{CEN} (since the ~ sign cannot be shown in LaTeX) and EC/CD = EN/ND = 1 as \frac{EC}{CD} = \frac{EN}{ND} = 1 (since this would like tidier) and nothing else. Don't worry, I did another formatting and it should be clean now. May 28 '15 at 17:21
• Looks like an intermediate editor meant to move a $, but forgot to delete it where it was to be moved from, and so added an additional $, making the markup all wrong. May 28 '15 at 17:46