Question

based on the information provided below, which two statements are true?
in △dab, m∠d = 30° and m∠dab = 120°. in △abc, (overline{ab} = 10) and (overline{bc} = 10).
(overline{dc} = 20)
(overline{db} = overline{dc})
△dbc is a right triangle
△dbc is an isosceles triangle
(2(da) = db)

Explanation:

Step1: Analyze △DAB

In △DAB, \( m\angle D = 30^\circ \), \( m\angle DAB = 120^\circ \), so \( m\angle DBA=180 - 30 - 120 = 30^\circ \). Thus, \( \triangle DAB \) is isosceles with \( DA = AB = 10 \), so \( DB = 2 \times AB \times \cos(30^\circ) \)? Wait, no, using the Law of Sines: \( \frac{AB}{\sin 30^\circ}=\frac{DB}{\sin 120^\circ} \). \( AB = 10 \), \( \sin 30^\circ=\frac{1}{2} \), \( \sin 120^\circ=\frac{\sqrt{3}}{2} \). So \( \frac{10}{\frac{1}{2}}=\frac{DB}{\frac{\sqrt{3}}{2}} \), \( DB = 10\sqrt{3} \)? Wait, maybe better to see \( \angle D = \angle DBA = 30^\circ \), so \( DA = AB = 10 \), so \( DC = DA + AC \). In \( \triangle ABC \), \( AB = BC = 10 \), so \( \triangle ABC \) is isosceles. \( \angle BAC = 180 - 120 = 60^\circ \) (since \( \angle DAB = 120^\circ \), linear pair), so \( \triangle ABC \) is equilateral (AB=BC=10, \( \angle BAC = 60^\circ \)), so \( AC = 10 \), thus \( DC = DA + AC = 10 + 10 = 20 \). So first statement \( \overline{DC}=20 \) is true.

Step2: Analyze \( \overline{DB}=\overline{DC} \)

From above, \( DC = 20 \), \( DB \): in \( \triangle DAB \), using Law of Cosines: \( DB^2 = DA^2 + AB^2 - 2 \times DA \times AB \times \cos(120^\circ) \). \( DA = AB = 10 \), \( \cos(120^\circ)=-\frac{1}{2} \). So \( DB^2 = 10^2 + 10^2 - 2 \times 10 \times 10 \times (-\frac{1}{2}) = 100 + 100 + 100 = 300 \), so \( DB = 10\sqrt{3} \approx 17.32
eq 20 \), so \( DB
eq DC \), second statement false.

Step3: Analyze \( \triangle DBC \) is right triangle

Check angles. \( DC = 20 \), \( BC = 10 \), \( DB = 10\sqrt{3} \). Check \( BC^2 + DB^2 = 100 + 300 = 400 = 20^2 = DC^2 \). So by Pythagorean theorem, \( \triangle DBC \) is right triangle (right-angled at B). So third statement true.

Step4: Analyze \( \triangle DBC \) is isosceles

\( DB = 10\sqrt{3} \), \( DC = 20 \), \( BC = 10 \). All sides different, so not isosceles. Fourth statement false.

Step5: Analyze \( 2(DA)=DB \)

\( DA = 10 \), \( 2(DA)=20 \), \( DB = 10\sqrt{3} \approx 17.32
eq 20 \), so false.

So true statements are \( \overline{DC}=20 \) and \( \triangle DBC \) is right triangle.

Answer:

\(\overline{DC} = 20\), \(\triangle DBC\) is a right triangle (the first and third options, assuming the options are: \( \overline{DC} = 20 \), \( \overline{DB} = \overline{DC} \), \( \triangle DBC \) is a right triangle, \( \triangle DBC \) is an isosceles triangle, \( 2(DA) = DB \))