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Isosceles Trapezoid

How to use
  1. ▶ Play runs the proof, Next → and ← Back step one scene at a time, Reset restores the opening scene and the original figure. Learn more about the controls
  2. Drag DD to reshape the trapezoid (CC follows), or AA or BB; the readout under the canvas shows the two base angles and the two diagonals. Learn more about why it holds
  3. Labels toggles the point names, Measures writes each marked angle's size beside its arc. The Step explanations log keeps every step so far. Learn more about reading a scene
Proof?AB ∥ DC and the legs AD = BC. Through C draw the parallel to AD, meeting AB at E: AECD is a parallelogram, so CE = AD = BC and the triangle EBC is isosceles with equal base angles at E and B. The angle at E equals the angle at A (corresponding angles on the parallels CE and DA), so ∠A = ∠B. Then SAS on the triangles DAB and CBA gives AC = BD. Drag D (C follows symmetrically), or A or B.Isosceles trapezoid
Display
Drag D to reshape the trapezoid (C follows), or A or B
ABDC
∠A = 63°, ∠B = 63°, AC = 5.09, BD = 5.09
Step 1 / 6

Step explanations

1A trapezoid with equal legs
Given
ABCD is a trapezoid: AB ∥ DC (the chevrons), and its legs are equal, AD = BC (the ticks). To prove: ∠DAB = ∠CBA (the base angles are equal) and AC = BD (the diagonals are equal).
Drag D: C follows and the legs stay equal. Learn more about the statement

Getting Started with the Visualizer
The Statement
A Parallel Through C
An Isosceles Triangle
Corresponding Angles
The Base Angles
Why It Holds
Dragging the Figure
Consequences



Getting Started with the Visualizer

DemoA run through the proof player
Step 0 of 7
Step through the proof, then drag DD to reshape the trapezoid (CC follows so the legs stay equal), or drag AA or BB, and watch the two base angles and the two diagonals in the readout stay equal. The picture below is the player with its parts numbered; the other tabs list what can be dragged, what each control does and what the step log shows.
Opening scene: the controls
1234567
1Proof — the theorem loaded in the player; the ? gives its idea in one line
2Display — Labels shows the point names, Measures writes each marked angle's size beside its arc
3Draggable points — D moves freely and C follows as its mirror image, so the legs stay equal; A and B move the long base; the hollow point E follows
4Readout — the line under the figure: the two base angles and the two diagonals, live while you drag
5Back, Play, Next, Reset — step through the proof or run it; Reset returns to step 1 and restores the original figure
6Speed and step counter — how fast Play advances, and which of the 6 steps is on screen
7Step explanations — every step reached so far, with its justification tag and a link to the section that explains it; click an entry to jump back to it
Proof complete: step 6 of 6
12345
1The parallelogram AECD — shaded: CE ∥ AD carries the leg over, CE = AD
2The isosceles triangle EBC — shaded: CE = CB, base angles β at E and B
3The diagonals — AC and BD, equal by SAS on DAB and CBA
4Justification tags — Given, Parallelogram, Isosceles base angles, Parallel lines, Combining, SAS
5Replay — at the last step Play becomes Replay and Next is off

The Statement

A trapezoid is a quadrilateral with one pair of parallel sides, the bases; the other two sides are the legs. When the legs are equal the trapezoid is isosceles, and two things follow:

∠DAB=∠CBA,AC=BD\angle DAB = \angle CBA, \qquad AC = BD


the angles at each base are equal, and the diagonals are equal. (The converse holds too: equal base angles, or equal diagonals, make the legs equal.) The player's trapezoid has AB∥DCAB \parallel DC and AD=BCAD = BC; DD is free, and CC is placed as its mirror image across the perpendicular bisector of ABAB, which is exactly what keeps the legs equal.

The proof is a chain of four earlier pages: a parallelogram, the isosceles base angles, parallel lines and SAS. The readout shows the two base angles and the two diagonals.
ABDC
Opening scene, frozen

The trapezoid: AB ∥ DC with chevrons, AD = BC with ticks.


A Parallel Through C

The trick is to move the leg ADAD next to the leg BCBC. Through CC draw the line parallel to ADAD; it meets the long base ABAB at EE. In the quadrilateral AECDAECD both pairs of opposite sides are parallel, AE∥DCAE \parallel DC (the bases) and EC∥ADEC \parallel AD (by construction), so it is a parallelogram and its opposite sides are equal:

CE=AD=BCCE = AD = BC


The two legs now meet at CC, as the two equal sides of the triangle EBCEBC.
ABDCE
A parallel through C, frozen

CE ∥ AD; AECD shaded as a parallelogram, so CE = AD = BC.


An Isosceles Triangle

The triangle EBCEBC has CE=CBCE = CB, so it is isosceles with apex CC, and its base angles are equal:

∠CEB=∠CBE=β\angle CEB = \angle CBE = \beta


The angle at BB is the base angle of the trapezoid itself; the angle at EE is a copy of it, waiting to be moved to AA.
ββABDCE
An isosceles triangle, frozen

EBC shaded: CE = CB, so the base angles at E and B are both β.


Corresponding Angles

The lines DADA and CECE are parallel, and the base ABAB crosses both. The angles they make with it on the same side are corresponding angles, and corresponding angles on parallel lines are equal:

∠DAB=∠CEB=β\angle DAB = \angle CEB = \beta
βββABDCE
Corresponding angles, frozen

β at A as well: DA ∥ CE with the transversal AB.


The Base Angles

Chain the two equalities: ∠DAB=∠CEB\angle DAB = \angle CEB and ∠CEB=∠CBA\angle CEB = \angle CBA, so

∠DAB=∠CBA=β\angle DAB = \angle CBA = \beta


The angles at the other base follow at once: ∠ADC\angle ADC and ∠DAB\angle DAB are co-interior angles between the parallels DCDC and ABAB, so ∠ADC=180°−β\angle ADC = 180° - \beta, and likewise ∠BCD=180°−β\angle BCD = 180° - \beta. Each base of an isosceles trapezoid carries a pair of equal angles.
βββABDCE
The base angles, frozen

∠DAB = ∠CBA = β.


Why It Holds

The diagonals come from one congruence. Compare the triangles DABDAB and CBACBA: they have AD=BCAD = BC (the legs), the base ABAB in common, and the included angles ∠DAB=∠CBA\angle DAB = \angle CBA just proved. By SAS they are congruent, and the sides opposite the equal angles are equal:

BD=ACBD = AC


The end-of-proof mark ∎ in the log closes the argument. The tags name the facts used: the given figure, a parallelogram, isosceles base angles, parallel lines, combining, and SAS.

The one idea worth remembering is the parallel through CC: it turns the two legs into the two equal sides of one triangle, and from there everything is a theorem already proved.
βββABDCE
The conclusion, frozen

The diagonals drawn: DAB ≅ CBA by SAS, so AC = BD.


Dragging the Figure

Drag DD: the top base slides and tilts, CC moves as the mirror image of DD so that the legs stay equal, the point EE and the parallel through CC follow, and in the readout the two base angles stay equal and the two diagonals stay equal. Drag AA or BB to change the long base; the figure stays symmetric about its perpendicular bisector.

The player keeps DD above ABAB and its foot between AA and the midpoint of ABAB, so that DCDC is the shorter base and EE falls inside ABAB as in the proof. A drag that would carry a point out of the frame, flatten the trapezoid or push EE past BB is refused.
βββABDCE
After dragging D, frozen

D moved up and to the right: steeper legs and a shorter top base, the base angles and diagonals still equal.


Consequences

• The converses. A trapezoid with equal base angles is isosceles (the same parallel through CC makes EBCEBC isosceles the other way round), and so is a trapezoid with equal diagonals (SSS on DABDAB and CBACBA gives equal base angles first).
• Cyclic. An isosceles trapezoid is the only trapezoid that is cyclic: its opposite angles are β\beta and 180°−β180° - \beta, which add up to 180°180°. Every cyclic trapezoid is isosceles.
• An axis of symmetry. The perpendicular bisector of the bases is a line of symmetry; the diagonals cross on it, and the midsegment is perpendicular to it.
• Rectangles. A rectangle is an isosceles trapezoid with both pairs of sides parallel; the equal diagonals of a rectangle are the special case β=90°\beta = 90°.
• Regular polygons. Any four consecutive vertices of a regular polygon form an isosceles trapezoid, which is why the diagonals of a regular pentagon are all equal.