How do you prove the Rhombus Diagonals Theorem?

Answer 1

ABCD is a rhombus if and only if its diagonals bisect each other and are perpendicular.

Proof: suppose ABCD is a rhombus with diagonals AC and BD. Call their point of intersection P. By the definition of a rhombus, AB≅BC≅CD≅DA. Clearly, by reflexivity, AC≅AC. Therefore, ABC≅CDA by SSS, so in particular ∠BAC≅∠DCA. But ∠BAC=∠BAP, and ∠DCA=∠DCP, so this means ∠BAP≅DCP. Also, ∠APB is vertical to ∠CPD, and therefore ∠APB≅∠CPD. This means APB≅CPD by AAS, so AP≅CP, which means BD bisects AC. Also, from the same congruence, we have that BP≅DP, so AC bisects BD. Now, BP≅BP by reflexivity, and since AP≅CP and AB≅BC, we have that ABP≅CBP by SSS. So in particular, ∠APB≅∠CPB. However, ∠APB and ∠CPB are supplementary, so m∠APB+m∠CPB = 180°, which implies that m∠APB+m∠APB = 180° → 2 m∠APB = 180° → m∠APB=90°, so AC is in fact perpendicular to BD.

For the other direction, suppose the diagonals bisect each other and are perpendicular. So AP≅CP and BP≅DP. Also, since AC and BD are perpendicular, we have m∠APB=m∠CPB=m∠CPD=m∠APD=90°, so ∠APB≅∠CPB≅∠CPD≅∠APD. Finally, we have that DP≅DP, CP≅CP, BP≅BP, and AP≅AP, all by reflexivity. This means APB≅CPB≅CPD≅APD, all by SAS. Therefore, by CPCTC, we have that AB≅CB≅CD≅AD, as required.

Note that one important detail of the first proof (namely, showing that segment AC actually intersects BD) has been omitted. I don't know whether you are expected to prove that or are simply allowed to assume it, and if you are expected to provide a proof, I don't know what plane separation axioms you are allowed to use. If you have to give a proof of that, you should try to prove that for AB and AD to be distinct line segments, D is on the opposite side of AC from B, and likewise for AB and CB to be distinct A is on the opposite side of BD from C. So segment BD intersects line AC (since B and D are on opposite sides), and it intersects line AC between A and C (or else A and C would be on the same side of BD), so in fact it intersects segment AC.

Answer 2

you know that a^2 + b^2 = c^2 regarding right triangles, and that the definition of a rhombus is four equal sides.

-so i can express this idea correctly draw the rhombus (in your head or on paper) as a dimond so that it has a vertical diagonal and a horizontal diagonal-

so you draw the diagonals and you have created four right triangles out of the rhombus.

so each of the triangles has a hypotenuse (c) of the same length.

also each triangle has half of the vertical diagonal as one side (a) and half of the horizontal diagonal as the other (b).

if a^2 = b^2 then a = b the combined quantity of half of a vertical and half a horizontal is always equal to the combined quantity of the other halves