Chapter 8.4 — Exercise 8.3 — Diagonals of Parallelogram

Bisecting Diagonals, Both Ways

In parallelogram ABCD, diagonals AC and BD meet at O. In △AOB and △COD: AB = CD (opposite sides of a parallelogram), ∠OAB = ∠OCD (alternate angles, since AB ∥ DC), and ∠OBA = ∠ODC (alternate angles, reading BD as a transversal across the same parallel pair AB ∥ DC) — ASA, so △AOB ≅ △COD, and CPCT gives OA = OC and OB = OD: the diagonals bisect each other.

The reverse statement is just as important as the theorem itself: if a quadrilateral's diagonals are already known to bisect each other, that quadrilateral must be a parallelogram, with no need to check either pair of sides for parallelism directly. In quadrilateral ABCD with OA = OC and OB = OD, △AOB ≅ △COD by SAS (OA = OC, ∠AOB = ∠COD as vertically opposite angles, OB = OD), giving ∠OAB = ∠OCD by CPCT — alternate angles equal, so AB ∥ DC. The identical argument applied to △AOD and △COB gives AD ∥ BC. Both pairs of opposite sides parallel, so ABCD is a parallelogram. This converse becomes one of the fastest ways to prove a figure is a parallelogram in the problems below, since it only ever needs one measurement fact about each diagonal rather than anything about angles or side lengths directly.

Solving for the Angles Algebraically

Parallelogram ABCD has opposite angles ∠A = (3x − 2)° and ∠C = (x + 48)°. Since opposite angles of a parallelogram are equal, 3x − 2 = x + 48, giving 2x = 50 and x = 25 — so ∠A = ∠C = 3(25) − 2 = 73°. Since adjacent angles are supplementary, ∠B = 180° − 73° = 107°, and ∠D = ∠B = 107° by the same opposite-angles rule. The four angles are 73°, 107°, 73°, 107° in order around the figure.

A closely related problem swaps the setup slightly: one angle is 24° less than twice the smallest angle. Calling the smallest angle x°, the other angle is (2x − 24)°, and since these two are adjacent (hence supplementary), x + 2x − 24 = 180, giving 3x = 204 and x = 68° — so the four angles work out to 68°, 112°, 68°, 112°. Both problems reduce to the identical two-step process: use "opposite angles equal" or "adjacent angles supplementary" to write one linear equation in a single unknown, solve it, then read every other angle off that one solved value using the same two rules again.

Extending a Side Until It Doubles

In parallelogram ABCD, E is the midpoint of BC; DE is extended to meet AB extended at F, and the goal is to show AF = 2AB. In △CED and △BEF: ∠CED = ∠BEF (vertically opposite angles), ∠DCE = ∠FBE (alternate angles, since DC ∥ AB), and CE = BE (E is the midpoint of BC) — AAS, so △CED ≅ △BEF, and CPCT gives CD = BF. Now AF = AB + BF = AB + CD (substituting), and since CD = AB in any parallelogram, AF = AB + AB = 2AB.

The midpoint condition is what makes AAS available here rather than needing a fourth fact: without E specifically bisecting BC, CE and BE wouldn't be guaranteed equal, and the whole congruence would collapse. This is a recurring shape in coordinate and synthetic geometry alike — extending a cevian from a midpoint through a vertex until it meets an extended side reliably doubles a length, precisely because the midpoint hands over one exact equal-segments fact for free.

An Exterior Bisector Builds a New Parallelogram

Isosceles triangle ABC has AB = AC; AD bisects the exterior angle at A (∠QAC, where BA is extended to Q) and CD ∥ BA. The goal is to show first ∠DAC = ∠BCA, then that ABCD is a parallelogram. Since AB = AC, ∠ABC = ∠BCA. The exterior angle ∠QAC equals the sum of the two remote interior angles, ∠ABC + ∠BCA = 2∠BCA (using the isosceles equality just noted). Since AD bisects ∠QAC, ∠DAC = ½∠QAC = ½(2∠BCA) = ∠BCA — the first part.

Since ∠DAC = ∠BCA and these are alternate angles across the transversal AC, AD ∥ BC. Combined with CD ∥ BA (given directly), both pairs of opposite sides of ABCD are now parallel, so ABCD is a parallelogram — this time proved through the original defining condition rather than the diagonal-bisection converse used elsewhere in this exercise.

Trisecting a Diagonal With the Converse Theorem

In parallelogram ABCD, diagonals AC and BD meet at O; P and Q are chosen on BD so that BP = PQ = QD (three equal pieces — the points of trisection). The goal is to prove AP ∥ CQ, and that AC bisects PQ. Since O is the midpoint of BD, OB = OD; subtracting the equal pieces BP and QD from each side, OP = OQ. Combined with the already-known OA = OC (diagonals of parallelogram ABCD bisect each other), quadrilateral APCQ has both its own diagonals, AC and PQ, bisecting each other at O — so by the converse theorem proved above, APCQ is itself a parallelogram, which immediately gives AP ∥ CQ (opposite sides of that new parallelogram) and confirms AC bisects PQ at O (since O is already established as PQ's midpoint).

This problem is really the converse theorem reused one level up: instead of applying it to the original parallelogram ABCD, it's applied to a second, smaller parallelogram, APCQ, built from points chosen inside the first one — a good illustration of how a proved theorem becomes a reusable tool rather than a one-time result tied to a single figure.

A Square's Midpoints Build Another Square

Square ABCD has E, F, G, H marking points on AB, BC, CD, DA respectively, with AE = BF = CG = DH. The goal is to show EFGH is itself a square. Since AE = BF = CG = DH and each full side of the square is equal, the remaining pieces are equal too: BE = CF = DG = AH. In the four corner triangles △AEH, △BFE, △CGF, △DHG: the two legs around each right angle match (AE = BF = CG = DH and AH = BE = CF = DG) and every corner angle is 90° (a square's own angle) — SAS, so all four corner triangles are congruent, giving EH = FE = GF = HG: all four sides of EFGH equal, making it at least a rhombus. Since the corner triangles are also isosceles (each has two equal legs), their base angles are 45° each, so the angle at E inside EFGH is 180° − 45° − 45° = 90° — and by the same reasoning at every other vertex, EFGH has four right angles too, which combined with four equal sides makes it a square.

From Diagonals to Midpoints

Every proof in this exercise has revolved around a parallelogram's diagonals, either directly or through a second parallelogram built inside the first. Exercise 8.4 shifts the focus one more time, to what happens when midpoints of a triangle's or a quadrilateral's sides are joined together instead of its diagonals.