It is a proof.
Given: line DB bisects line AC
line AD is parallel to line BE
AD=BE
Prove: DB=EC
there are two triangles connected together by point B. They are labled A D B and B E C. D and E are the top points of the triangles. they look like they would be right angle triangles but it isn't proven. B and C would be the right angles they are facing the same way
Suggestions: use SAS to show the triangles are congruent. Since DB bisects AC, AB=AC. since angle DAB = angle EBC (why?), triangle ADB is oongruent to triangle BEC.
Hint for the angles: AD || BE, so AC is a transversal. Now look for something about corresponding angles.
To prove that DB = EC, we need to show that triangle ADB is congruent to triangle BEC using the Side-Angle-Side (SAS) congruence criterion.
Given that line DB bisects line AC, we know that AB = AC (since a line bisects another line when it divides it into two equal parts).
We are also given that AD is parallel to BE, which means that angle DAB is equal to angle EBC. This is because corresponding angles formed by a transversal (in this case, AC) and two parallel lines (AD and BE) are congruent.
Now, let's apply the SAS congruence criterion to triangles ADB and BEC:
- Side AB = Side EC (since AB = AC and AC = EC from the information given)
- Angle DAB = Angle EBC (as explained above)
- Side AD = Side BE (given AD = BE)
Since both triangles have two corresponding sides and angles that are congruent, we can conclude that triangle ADB is congruent to triangle BEC by the SAS criterion.
By definition of congruence, corresponding parts of congruent triangles are congruent. Therefore, DB = EC, as required to be proven.