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Given:
m/_DAC=m/_DBC;

bar(AB) bisects
/_DAC;

bar(AB) bisects
/_DBC.
Prove:
m/_1=m/_2

Given: $m \angle D A C=m \angle D B C$ ; $\newline$ $\overline{A B}$ bisects $\angle D A C$ ; $\newline$ $\overline{A B}$ bisects $\angle D B C$ . $\newline$ Prove: $m \angle 1=m \angle 2$

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Find derivatives using the product rule II

Full solution

Q. Given:

m \angle D A C=m \angle D B C

;

\newline

\overline{A B}

bisects

\angle D A C

;

\newline

\overline{A B}

bisects

\angle D B C

.

\newline

Prove:

m \angle 1=m \angle 2

Identify Given Information: Let's identify the given information and what we need to prove. $\newline$ Given: $\newline$ $1$ . $m/\_DAC = m/\_DBC$ (The measure of angle DAC is equal to the measure of angle DBC). $\newline$ $2$ . $\overline{AB}$ bisects $/\_DAC$ (Line segment AB bisects angle DAC). $\newline$ $3$ . $\overline{AB}$ bisects $/\_DBC$ (Line segment AB bisects angle DBC). $\newline$ We need to prove that $m/\_1 = m/\_2$ .
Angle Division by Line Segment: Since $\overline{AB}$ bisects $\angle DAC$ , it divides angle DAC into two equal angles. Let's denote the measure of these two angles as $m\angle_1$ . Therefore, we have: $\newline$ $m\angle_{DAC} = 2 \times m\angle_1$ .
Measure of Angles: Similarly, since $\bar{AB}$ bisects $/_DBC$ , it divides angle DBC into two equal angles. Let's denote the measure of these two angles as $m/_2$ . Therefore, we have: $\newline$ $m/_DBC = 2 \times m/_2$ .
Substitution of Measures: From the given information, we know that $m/_{DAC} = m/_{DBC}$ . By substituting the expressions from the previous steps, we get: $\newline$ $2 \times m/_{1} = 2 \times m/_{2}.$
Final Measure Calculation: To find the measure of each angle, we can divide both sides of the equation by $2$ , which gives us: $\newline$ $m/_{1} = m/_{2}.$

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