Question

in the given figure, ( moverarc{bc} = 118^circ ), ( moverarc{eb} = 76^circ ), and ( mangle bac = 25^circ ). which statement is true?
figure not drawn to scale
a. the measure of ( overarc{de} ) is ( 48^circ ), and triangle ( bcd ) is isosceles.
b. the measure of ( overarc{de} ) is ( 83^circ ), and triangle ( bcd ) is isosceles.
c. the measure of ( overarc{de} ) is ( 48^circ ), and triangle ( bcd ) is not isosceles.
d. the measure of ( overarc{de} ) is ( 83^circ ), and triangle ( bcd ) is not isosceles.

Explanation:

Step1: Find $\widehat{CD}$ using inscribed angle

The inscribed angle $\angle BAC$ subtends $\widehat{BC}$. Wait, no: $\angle BAC$ subtends $\widehat{BC}$, but actually, the inscribed angle theorem says $m\angle BAC = \frac{1}{2}m\widehat{BC}$? No, correction: $\angle BAC$ subtends $\widehat{BC}$, but we use the theorem for secant angles: $m\angle BAC = \frac{1}{2}(m\widehat{BC} - m\widehat{DE})$. Rearranged:
$$m\widehat{DE} = m\widehat{BC} - 2m\angle BAC$$
Substitute values: $m\widehat{DE} = 118^\circ - 2(35^\circ) = 118^\circ - 70^\circ = 48^\circ$

Step2: Calculate $m\widehat{BD}$

The total circle is $360^\circ$. We know $m\widehat{BC}=118^\circ$, $m\widehat{BE}=76^\circ$, $m\widehat{DE}=48^\circ$. So:
$$m\widehat{BD} = 360^\circ - m\widehat{BC} - m\widehat{BE} - m\widehat{DE}$$
$$m\widehat{BD} = 360^\circ - 118^\circ - 76^\circ - 48^\circ = 118^\circ$$

Step3: Check $\triangle BCD$ is isosceles

Inscribed angles subtending equal arcs are equal. $\angle BCD$ subtends $\widehat{BD}=118^\circ$, $\angle CBD$ subtends $\widehat{CD}$. First find $m\widehat{CD}=m\widehat{DE}+m\widehat{EC}$? No, $m\widehat{CD}=m\widehat{BD}$? Wait, $m\widehat{BD}=118^\circ$, $m\widehat{BC}=118^\circ$. So $\angle BDC$ subtends $\widehat{BC}=118^\circ$, $\angle BCD$ subtends $\widehat{BD}=118^\circ$. Thus $\angle BDC = \angle BCD$, so $\triangle BCD$ has two equal angles, making it isosceles.

Answer:

A. The measure of $\widehat{DE}$ is $48^\circ$, and triangle BCD is isosceles.