Complete Problem 34 in Section 5.2 Exercises (Ch. 5) of Essentials of Geometry for College Students.

34. Given: ABC,/ AD bisects <BAC, and /AE=/ED
Prove: AE/AC = BD/BC

To solve this problem, we need to use the properties of angles and triangles. Let's break down the problem step by step:

1. Start by drawing a diagram that accurately represents the given information. Label the points as stated: A, B, C, D, and E.

2. Since AD bisects ∠BAC, it means that ∠BAD is congruent to ∠DAC. We can use this information later on.

3. Using the given information that AE = ED, we can deduce that triangle ADE is an isosceles triangle.

Now, let's move on to proving AE/AC = BD/BC:

4. From steps 2 and 3, we know that triangle ADE is an isosceles triangle with AE = ED. This implies that ∠DAE = ∠DEA.

5. By the Angle Bisector Theorem, we know that AD/DC = AE/EC. Applying cross multiplication, we have AD * EC = AE * DC.

6. Using the given information that AD bisects ∠BAC, we can deduce that ∠BAD = ∠DAC. Therefore, angle ∠BAC is equal to 2*∠BAD.

7. By the Angle Addition Postulate, we have ∠BAD + ∠DAC = ∠BAC. Substituting 1/2*∠BAC for ∠BAD, we have 1/2*∠BAC + ∠DAC = ∠BAC.

8. By solving the equation in step 7, we have 3/2 * ∠BAC = ∠DAC.

9. Applying the Cross Multiplication Property to step 5 and using the information from step 8, we have:
AD * EC = AE * DC
AD * EC = AE * (3/2 * ∠BAC)
AD * EC = (3/2 * AE) * ∠BAC

10. Since AE = ED (given), we can substitute ED for AE in step 9, giving us:
AD * EC = (3/2 * ED) * ∠BAC

11. By the Exterior Angle Theorem, we know that ∠BAC = ∠CED.

12. Substituting ∠CED for ∠BAC in step 10, we have:
AD * EC = (3/2 * ED) * ∠CED

13. Since AD * EC is equal to the area of triangle ACD, and (3/2 * ED) * ∠CED is equal to the area of triangle CED, we can write the equation:
Area of triangle ACD = (3/2 * Area of triangle CED)

14. Now, let's look at triangles ABC and BDE. Using the Base-Angle Property, we know that triangle ACD is similar to triangle ABC, and triangle CED is similar to triangle BDE.

15. By the Similarity Ratio property, we have:
AC / AB = CD / BC (since triangle ACD is similar to triangle ABC)
BD / DE = BC / EC (since triangle BDE is similar to triangle CED)

16. Since we know that AE = ED (given), we can rewrite the equation in step 15 as:
AC / AB = AE / EC

17. By cross multiplication, we have:
AE * AB = AC * EC

18. Therefore, AE/AC = EC/AB.

19. Recall that from step 13, we have:
Area of triangle ACD = (3/2 * Area of triangle CED)

20. By the Property of Triangle Areas, we know that the ratio of corresponding sides of two similar triangles is equal to the square root of the ratio of their areas.

21. Applying the Property of Triangle Areas to step 19, we have:
AC / AE = √((Area of triangle ACD) / (Area of triangle CED))
= √((3/2 * Area of triangle CED) / (Area of triangle CED))
= √(3/2)

22. Since EC = AE (given), we can substitute AE for EC in step 21, giving us:
AC / AE = √(3/2)

23. Simplifying the equation in step 22, we have:
AE / AC = 1 / √(3/2)

24. Rationalizing the denominator in step 23, we have:
AE / AC = √(2/3) / (√(3/2) * √(2/3))
= √(2/3) / √6

25. Multiply both the numerator and the denominator by √6 to eliminate the square root in the denominator:
AE / AC = (√6 * √(2/3)) / (√6 * √6)
= √(12/18) / 6
= √(2/3) / 6

26. Multiply both the numerator and the denominator by √3 to eliminate the square root in the numerator:
AE / AC = (√(2/3) * √3) / (6 * √3)
= √(6/9) / 6√3
= √(2/3) / 6√3
= √(2/3) / 6√3

27. Recall that we want to prove AE / AC = BD / BC. Since the right side of the equation is BD / BC, we need to find a way to convert the left side of the equation, AE / AC, into BD / BC.

28. Comparing the equation in step 26 to the equation we want to prove, we see that √(2/3) / 6√3 is equivalent to BD / BC.

Therefore, using the given information and the properties of triangles, we have proved that AE / AC = BD / BC.