As Archimedes merely states the theorem, the proof does not pretend to be a reconstruction.

We use the notation of Dijksterhuis for the square on a straight-line and rectangle formed from two straight lines. T(AB) is the square on straight-line AB, while O(AB,CD) is the rectangle formed from straight-lines AB and CD (or lines equal to AB and CD).

Theorem 1: If there is an orthotome, ABG, and BD is parallel to its diameter or itself a diameter,  and AG is parallel to the tangent to the orthotome at B, then AD will be equal to DG. And if AD is equal to DG, then AG and the tangent to the orthotome at B will be parallels.
Case where BD is parallel to the diameter. Where BD is the diameter go to Theorem 0B. This proof depends on all of Theorem 0.

(Basic Diagram for first part)
 (diagram1) Let ABG be an orthotome and BD parallel to the diameter EZ of the orthotome, I say that AD = DG. (Here we give only one case, where ABG does not intersect ZE). (diagram 2) Let tangent BH at B be drawn (by Theorem 0D) and QE extended to H. Let perpendiculars, BQ, AZ, GK be drawn. These are ordinates of the parabola for the axis EZ (diagram 3 = basic ) Let AG be extended to meet HZ at L, BD be extended to meet AZ at M, and let DN perpendicular to HZ be drawn, and let GRST be drawn parallel to AH to meet AZ at T. Since AZ, DN, BQ, and GK are all perpendicular to HZ, they are clearly parallel, and since HZ is parallel to GT, they are also perpendicular to GT. (Basic Diagram for second part) (diagram 4) Since triangles, AZL, BQH, and GKL are similar: AZ : BQ : GK = ZL : QH : KL Hence, T(AZ) : T(BQ) : T(GK) = T(ZL) : T(QH) : T(KL). (diagram 5) However, by Theorem 0C, T(AZ) : T(BQ) : T(GK) = ZE : QE : KE. Hence, T(ZL) : T(QH) : T(KL) = ZE : QE : KE. (diagram 6) But QE = 1/2 QH, so that T(ZL) : T(QH) : T(KL) = ZE : 1/2*QH : KE. Taking the first and third ratios, T(ZL) : T(KL) = ZE : KE, it follows separando that T(ZL) - T(KL) : T(KL) = ZK : KE. In other words, O(ZL+KL, ZK) : T(KL) = ZK : KE. Taking also the second and third ratios, T(QH) : T(KL) = 1/2 QH : KE, it follows ex aequali that O(ZL+KL, ZK) : T(QH) = ZK : 1/2 QH. Or: O(ZL+KL, ZK) : T(QH) = 2*ZK : QH Or: O(ZL+KL, ZK) : T(QH) = O(2*ZK,QH) : T(QH). Hence, O(ZL+KL, ZK) = O(ZK,2*QH), Whence, ZL+KL= 2*QH. (diagram 7) But, since DBQN and DBHL are parallelograms, NO = DB = HL, so that NL = QH. (diagram 8) Thus ZL + KL = 2*NL = 2*NK + 2*KL. But ZL = ZK + LK. Hence, ZK + 2*KL = 2*NK + 2*KL. That is, ZK = 2*NK. Thus, ZN = NK. Therefore, ZK = 2*NK. Hence, ZN = NK. (diagram 9) Since, ZN = MD, and NK = SG. (diagram 10) Therefore, the similar triangles AMD and DSG are equal. Therefore, AD = DG. (Basic Diagram for converse of the theorem) (diagram 11) Suppose that AD = DG, I say that AG is parallel to BH. (diagram 12) Suppose that they are not parallel. Join BF to ZH where BF is parallel to AG. We will prove that QE = EF. (diagram 13) Since triangles AMD and DSG are similar, they are also equal. Hence, ZN = MD = SQ = NK. And DB = LF = NQ. Hence QF = NL, or ZL = NZ + NL = NK + QF. (diagram 14) Therefore, AZ : BQ : GK = ZL : QF : KL = NK+QF : QF : KL Or: T(AZ) : T(BQ) : T(GK) = T(NK+QF) : T(QF) : T(KL). (diagram 15) Furthermore, by Theorem 0C, T(AZ) : T(BQ) : T(GK) = ZE : QE : KE. Hence, ZE : QE : KE = T(NK+QF) : T(QF) : T(KL) Taking the first and third ratios, ZE : KE = T(NK+QF) : T(KL), separando, ZK : KE = T(NK+QF) - T(KL) : T(KL) That is, 2*NK : KE = O(NK+QF-KL,NK+QF+KL) : T(KL) = O(NK+NL-KL,NK+NL+KL) : T(KL) = O(2*NK,2*NL) : T(KL) Or rather NK : KE = O(NK,2*OF) : T(KL), since NL = OF. Taking the second and third ratios, QE : KE = T(QF) : T(KL), ex aequali NK : QE = O(NK,2*OF) : T(QF) = O(NK,OF) : 1/2*T(QF) = NK : 1/2 OF. (diagram 16) Hence, QE = 1/2 QF, which is impossible, since it will then be tangent at B, contrary to the hypothesis.