Let ABCD be a convex quadrilateral with AD = BC. Show that AD and BC determine congruent angles with the line passing the midpoints.

Question

Let ABCD be a convex quadrilateral with AD = BC. Show that AD and BC determine congruent angles with the line passing through the midpoints of sides AB and CD...

MY IDEAS

MY DRAWING

As you can see i noted some points.

Okey, so, i thought of similar or congruent triangles. But i don't know where to start. Hope one of you cn help me! Thank you!

Answer 1

Extend $EF$ as necessary and mark the points $X$ and $Y$ on this line, so that $|AX|=|AE|$, $|BY|=|AF|$ (and $X$, $E$ are distinct unless $\angle AEF$ is a right angle, and a similar condition for $Y$, $F$). Then $ABYX$ is congruent to $BCFE$, so your result follows.

Answer 2

Another way of showing this would be as follows.

Reflect line $CB$ in line $EF$ to $C'B'$ and translate it parallel to $EF$ so that $C'$ lies on $D$ and $B'$ moves to $B''$, which is possible because the reflection will leave $C'D$ parallel to $EF$. Now $B''$ is the same distance from the line $EF$ as $A$ is, and the same distance from $D$ as $A$ is; so it is at $A$ and the result follows.

Answer 3

This is a pedant solution going into details. The picture below should make the idea of proof transparent. As in the question, let $G,H$ be the intersection of $EF$ with the side $BC$, respectively $AD$ of the given quadrilateral $ABCD$.

Let $A',B',C',D'$ be respectively the projections of $A,B,C,D$ on $EF$.

We may and do assume that, as in the picture, $AD\cap BC$ is a point in the closed half-plane $\mathcal H$ w.r.t. $AD$ not containing $B,C$. (Else we switch in the same time $A\leftrightarrow D$, and $B\leftrightarrow C$.

We may and do assume that $D'$ is in the same half-plane $\mathcal H$, and $C$ in the closed opposite half-plane.

Then we have:

• $AA'=BB'$ and $EA'=EB'$, since $E$ is the mid point of $AB$.
• $CC'=DD'$ and $FC'=FD''$, since $F$ is the mid point of $CD$.
• $(*)$ The points $B',E,A'$ come in this order on $EF$, and the points $C',F,D'$.
• This implies that $B'C'=A'D'$. We can now compare the two trapeziums $AA'D'D$ and $BB'C'C$. They have the same height $A'D'=B'C'$, right angles in $A',D';B',C'$, the same bases, $AA'=BB'$ and $DD'=CC'$, and the same remained corresponding side $AD=BC$. By drawing the heights from $D$ and $C$ in them, seen as scissors, the two trapeziums are cut into two equal rectangles with same height and same side $DD'=CC'$, and two triangles with one right angle each, same height from $D$ and $C$, and same opposite side to $D$ and $C$. So the two trapeziums are congruent. In particular, the angle between $AD$ and $A'D'$, which is $\hat H=\widehat{AHE}$, is the same one as the angle between $BC$ and $B'CD'$, which is $\hat G=\widehat{BGE}$.

$\square$

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Note: At some point above, marked with $(*)$, we have assumed that the points $B',E,A';C',F,D'$ come in a same order on the line $EF$. As "in the picture". To see that this fact is not an optical illusion, so that the proof does not depend on the picture, let us show this fact.

We start with the points $E,F$, with to-be-projections $A',B';C',D'$ on the line $EF$, so that $A'E=B'E$ and $C'F=D'F$, but so that the points are in an order not respecting $(*)$. Then we make a choice of $A,B,C,D$ on the perpendiculars on $EF$ in $A',B',C',D'$, so that $E$ is the mid point of $AB$, and $F$ of $CD$. In a picture:

Let $B''$, $C''$ be the reflections of $B,C$ w.r.t. $B',C'$. Alternatively, we can define them as the projections of $A,D$ on $BB'$, $CC'$.

If $(*)$ is not holding, then we have in principle the shown order of points on $EF$. (Or a similar one obtained by exchanging $AD$ with $BC$.)

Then we obtain a contradiction from $AD=BC=B''C''$ by using Pythagoras to compute the lengths of $AD$ and $B''C''$, using the projections of these segments on the direction $EF$, where the projections are $A'D'>B'C'$, and on the perpendicular direction, where the projections coincide.

Answer 4

The idea in this answer is similar to the one in dan_fulea's good answer.

Let $E,F$ be the midpoint of $AB,CD$ respectively.

Let $G$ be the intersection point of $EF$ with $BC$.

Let $H$ be the intersection point of $EF$ with $AD$.

We may suppose that we have the following :

• Let $I$ be a point on the line parallel to $EF$ passing through $D$ such that $AI\perp ID$.

• Let $J$ be a point on the line parallel to $EF$ passing through $C$ such that $CJ\perp JB$.

• $K,L,M,N$ be a point on $EF$ such that $KA\perp EF,CL\perp EF,DM\perp EF,BN\perp EF$ resepectively.

Then, we have $$DM=CL,\qquad AK=BN$$ since $\triangle{FMD}\equiv \triangle{FLC}$ and $\triangle{EKA}\equiv\triangle{ENB}$.

So, we have $$\triangle{AID}\equiv\triangle{BJC}$$ since $AD=BC$ and $$AI=AK-IK=BN-DM=BN-CL=BN-JN=BJ$$

So, we get $\angle{ADI}=\angle{BCJ}$.

Therefore, $\angle{AHE}=\angle{BGE}$ follows.