Kite Properties

Contents: Three Proofs | Comments: Another Case | Possible Mistakes | Use to Prove SSS

Given ABCD a kite, with AB = AD and CB = CD, the following things are true.

• Diagonal line AC is the perpendicular bisector of BD.
• The intersection E of line AC and line BD is the midpoint of BD.
• Angles AED, DEC, CED, BEA are right angles.
• Triangle ABC is congruent to triangle ADC.
• Consequently angle ABC = angle ADC.
• Line AC bisects angles BAD and BCD.

Two statements are in bold type, because those statements include the others, from the definitions or perpendicular bisector and congruence of triangles. (Of course to prove the bold statements, one may have to prove some of the others first.

Three Proofs found in Class

In class, students were asked to prove this:

Given a kite ABCD with AB = AD and CB = CD, then triangle ABC is congruent to triangle ADC.

Here are two proofs that were found in class (my wording).  (Note: this is an example that shows there is not necessarily just one good proof of a geometrical theorem.)

Proof 1:  (key idea: show angle BAC = angle DAC)

• Let M be the midpoint of BD.  Since triangle ABD is isosceles, ray AM bisects angle BAD, so angle BAM = angle DAM.
• Also, since triangle ABD is isosceles, line AM is perpendicular to BD. 
• Since also triangle CDB is isosceles, line CM is perpendicular to BD for the same reason.  Since there is only one line through M that is perpendicular to BD (protractor axiom) then line AM = line CM, so A, M, C are collinear and this line is line AC.
• Angle BAM = angle BAC and angle DAM = angle DAC (same rays).
• Thus triangle ABC is congruent to triangle ADC by SAS, since the two angles BAC and DAC above are congruent and also AB = AD (by hypothesis) and AC = AC.

QED (almost – see comments below about another case).

Proof 2: (key idea: show angle BAC = angle DAC)

• Let m be the perpendicular bisector of BD.  By definition of the perpendicular bisector, m intersects BD at M, the midpoint of BD.
• The triangle ABD is isosceles.  By the symmetry properties of the isosceles triangle, the line AM is the perpendicular bisector of BD = m.  Thus A is on m.
• Also, since triangle ABD is isosceles, ray AM bisects angle BAD, so angle BAM = angle DAM.
• Angle BAM = angle BAC and angle DAM = angle DAC (same rays)
• Thus triangle ABC is congruent to triangle ADC by SAS, since the two angles BAC and DAC above are congruent and also AB = AD (by hypothesis) and AC = AC.

QED (almost – see comments below about another case).

Proof 3: (key idea: show angle ABC = angle ADC)

• Triangle ABD is isosceles, so angle ABD = angle ADB.
• Triangle CBD is isosceles, so angle CBD = angle CDB.
• Angle ABC = angle ABD + angle DBC
• Angle ADC = angle ADB + angle BDC
• Since the sums on the right are equal, because of the congruent angles, then angle ABC = angle ADC.
• Thus triangle ABC is congruent to triangle ADC by SAS, since the two angles ABC = ADC are congruent and also AB = AD  and CB = CD (by hypothesis).

QED (almost – see comments below about another case).

Comments: Good proofs, but there is another (nonconvex) case

Each of the proofs is good, and valid for the figure shown at the top of the page.  However there is another figure that shows another case of a kite, for which the proofs are almost right, but need some additional comments.  This is this case where A and C are both on the same side of line BD, such as in this figure.

So what goes wrong with the proofs? 

In Proof 1 and Proof 2, an important step was

• Angle BAM = angle BAC and angle DAM = angle DAC (same rays).

But this is not true in this figure.  But all is not lost.  We just have to justify angle BAC = angle DAC in this case also. Then the proofs are correct for the remaining steps. We can actually establish and angle equality a couple of ways.

• It is still true that ray CM = ray CA, so we can change the roles of A and C and work with bisecting the angle at C instead of the angle at A.  We don’t have to write the whole proof again; we can just re-label ABCD so that A is the vertex farthest from M.
• A second approach is to work with the figure as labeled, but assert that two angles are equal if they are supplementary to equal angles.  In this case, angle BAC and angle BAM are supplementary, and also angle DAC is supplementary to angle DAM, which equals angle BAM. Thus angle BAC = angle DAC.

In Proof 3, the step that are not true in this care are

• Angle ABC = angle ABD + angle DBC
• Angle ADC = angle ADB + angle BDC

But what if we subtract instead of add, then we still get correct statements that can be used in the rest of the proof.

• Angle ABC = angle DBC - angle ABD
• Angle ADC = angle BDC - angle ADB

Possible Mistakes in Proofs

While there are a number of correct ways to prove the theorem above, there are also a number of erroneous approaches.  Typically, these include a disguised way of assuming the conclusion.  Here are some examples of beginnings of such proofs that triangle ABC is congruent to triangle ADC

Of course all these facts are true, but at the outset we are trying to prove the basic relations, so we can’t assume them to prove them.