Given: line AB is congruent to line AC, Angle BAD is congruent to angle CAD.
Prove: line AD bisects BC
Picture: An upside down triangle divided inhalf to form two triangles Angle BAD and angle CAD. They share a common side of AD. and point D bisects line BC. and Angle c and Angle B are corresponding and congruent.
what i have so far
given info
Line DA is congruent to Line DA by Reflexive Prop or EQuality
angle C is congruent to angle B by the Corresponding parts of congruent triangles are congruen. CPCTC.
but i do not know where to go from that..
As far as I can tell, it doesn't matter which lines are congruent.
<BAD and <CAD are adjacent angles which are equal.
Therefore they bisect <CAB
Gussy it up, but that's the crux of the argument.
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You know it's almost 2017 already.
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To prove that line AD bisects line BC, we can use the Angle Bisector Theorem. Here's how to proceed:
1. Given: Line AB is congruent to line AC (AB ≅ AC) and angle BAD is congruent to angle CAD (∠BAD ≅ ∠CAD).
2. We want to prove that line AD bisects line BC (BD ≅ CD).
To prove this, we will use the Angle Bisector Theorem, which states that in a triangle, an angle bisector divides the opposite side into segments that are proportional to the lengths of the other two sides.
3. Draw line segment BD and line segment CD.
4. Since AB ≅ AC, the triangle ABC is an isosceles triangle. Therefore, line segment BD ≅ CD (by the Converse of the Isosceles Triangle Theorem).
5. Now, consider triangle ABD and triangle ACD. These triangles share a common side (AD) and have congruent corresponding angles (∠BAD ≅ ∠CAD).
6. By the Angle-Angle (AA) similarity criterion, triangle ABD is similar to triangle ACD.
7. Since triangle ABD and triangle ACD are similar, their corresponding sides are proportional.
8. By the Angle Bisector Theorem, we know that the segments BD and CD are divided in proportion to the lengths of the other two sides, which means BD/CD = AB/AC.
9. Substituting the given congruence AB ≅ AC in the last equation, we have BD/CD = 1/1.
10. Thus, BD = CD, which proves that line AD bisects line BC.
Therefore, line AD bisects line BC, as desired.