In the diagram, given that $ADGB$ is an isosceles trapezoid, $CE = CD, BC = BD,$ $\angle AOD = 30^{\circ} $, prove that $BD = DF$.

Question

In the diagram, given that $ADGB$ is an isosceles trapezoid, $CE = CD, BC = BD,$ $\angle AOD = 30^{\circ} $,

please prove that $BD = DF$.

My idea: Or to prove $BD = DF$, only proving that $\angle DBO = 45^{\circ}$ would suffice.

Answer 1

B, A, D, G, F are con-cyclic points of the red circle (centered at O).

Construct the green dotted circle passing through E, D, F such that:-

(1) DF is the common chord of the two circles;

(2) OM is the line of centers (where M is the midpoint of DF);

(3) $\angle EDF = 90^0$; and

(4) CE = CD.

All the above can be used to pin point AECF is a straight line with C as the center of the green dotted circle; EF as its diameter; and CE = CD = CF = its radius.

Since AECF is a straight line, $\angle 2 = \angle 1 = 0.5 \angle AOD = 15^0$

$\theta_3 = \theta$ [angles in the same segment]

$= \theta_1$ [$AD//BG$]

$= \theta_2$ [because of all the brown-marked angles.]

$= 2 \times \angle 2 = 30^0$ [angle at center = 2 times angle at circumference]

Therefore, $\angle ODF = \angle OFD = … = 45^0$