Given: ABC,/ AD bisects <BAC, and /AE=/ED

Prove: AE/AC = BD/BC

To prove that AE/AC = BD/BC, we need to use the given information that AD bisects ∠BAC and AE/ED = 1.

Let's break down the proof step by step:

Step 1: Given information:
AD bisects ∠BAC and AE/ED = 1.

Step 2: Since AD bisects ∠BAC, it means that ∠BAD is congruent to ∠CAD. We can denote this as ∠BAD ≅ ∠CAD.

Step 3: Now, we can use the Angle Bisector Theorem, which states that if a line segment bisects an angle, then it divides the opposite side into two segments that are proportional to the adjacent sides. In our case, we can write:

AE/EC = AD/DC

Step 4: Since AE/ED = 1, we can substitute AE for ED in the equation from Step 3:

AE/EC = AD/DC = 1

Step 5: Since AE/EC = 1, we can multiply both sides of the equation by EC to simplify it:

AE = EC

Step 6: Now, let's look at the triangles AEC and BDC. We have two pairs of corresponding sides that are congruent: AE = EC and BD = DC.

Step 7: Using the Side-Splitter Theorem, we know that if a line intersects two sides of a triangle proportionally, then it divides the third side proportionally as well. In this case, we can write:

AE/AC = BD/BC

Step 8: Since we know from Step 5 that AE = EC, we can substitute EC for AE in the equation from Step 7:

EC/AC = BD/BC

Step 9: Simplifying the equation, we can write:

1/AC = BD/BC

Step 10: Multiplying both sides of the equation by AC, we get:

AC * (1/AC) = (BD/BC) * AC

Simplifying further, we have:

1 = (BD/BC) * AC

Step 11: Dividing both sides of the equation by 1, we get:

BD/BC = AC

Therefore, we have proved that AE/AC = BD/BC.