Given: m∥n , m∠1=65∘ , m∠2=60∘ , and BD−→− bisects ∠ABC . Prove: m∠6=70∘ Line m parallel to line n. Line t passing through both lines. There are four angles formed by lines m and t intersecting at point B. The lower left angle is labeled 5. The upper right angle is angle A B C with point A on line t and point C on line m. Angle A B C is separated by ray B D. The angle on the right side of the ray is labeled 4. The angle on the left side is labeled 3. A segment joins points A and D forming a triangle with angle 3 as one of its interior angles. The other two interior angles are labeled 1 and 2. There are four angles formed by lines n and t intersecting. The upper left angle is labeled 6. Put responses in the correct input to answer the question. Select a response, navigate to the desired input and insert the response. Responses can be selected and inserted using the space bar, enter key, left mouse button or touchpad. Responses can also be moved by dragging with a mouse. It is given that m∥n , m∠1=65∘ , m∠2=60∘ , and BD−→− bisects ∠ABC . Because of the triangle sum theorem, m∠3=55∘ . By the Response area, ∠3≅∠4 , so m∠4=55∘ . Using the Response area, m∠ABC=110∘ . m∠5=110∘ because vertical angles are congruent. Because of the Response area, m∠5+m∠6=180∘ . Substituting gives 110∘+m∠6=180∘ . So, by the Response area, m∠6=70∘ .