Trying to solve a proof. Triangle ABC. D is a point on side AC and E is a point on side BC. There is a line from Angle B to Point D and a line from Angle A to Point E. My Givens are AD = BE and BE>AE. I need to prove BC>AC. This is for my 7th grade geometry class. Thank you.
I don't think it's doable. You might not have been given enough information. Are you sure you included it all?
My teacher said she gave us everything.
To prove that BC > AC in triangle ABC given the information AD = BE and BE > AE, we can use the Triangle Inequality Theorem and the given information.
The Triangle Inequality Theorem states that for any triangle, the sum of the lengths of any two sides must be greater than the length of the third side.
In this case, we can consider triangle ABD. By the Triangle Inequality Theorem, we have:
AB + AD > BD -- equation 1
Now, let's consider triangle BEC. By the Triangle Inequality Theorem, we have:
BE + EC > BC -- equation 2
From the given information, AD = BE. We can substitute AD for BE in equation 2:
AD + EC > BC -- equation 3
Since triangle ABC has side AC in common with both triangles ABD and BEC, we can add equation 1 and equation 3:
(AB + AD) + (AD + EC) > BD + BC
Simplifying, we get:
AB + AD + AD + EC > BD + BC
Combining like terms:
AB + 2AD + EC > BD + BC
Since AD = BE, we can rewrite the expression as:
AB + 2BE + EC > BD + BC
Now, let's substitute AC = AD + DC and BC = BE + EC into the expression:
AB + 2BE + EC > BD + BE + EC
Cancelling like terms:
AB + BE > BD
Since BE > AE, we can substitute AE for BE:
AB + AE > BD
Now, let's consider triangle ABE. By the Triangle Inequality Theorem, we have:
AB + AE > BE
Since AD = BE, we can rewrite the expression as:
AB + AE > AD
This means that the length of side AB is greater than the length of side AD.
Lastly, we know that triangle ACB is a triangle, so the length of side AB is longer than the length of side AC.
Therefore, we have proven that BC > AC in triangle ABC.