AD is an altitude of an isosceles triangle ABC in which AB=AC. Show that AD bisects BC.

The two halves of the triangle are congruent. (Proof: SAS. Side, bisected half angle A/2 and common side are equal)

Therefore the two lengths BD and DB are equal and BC is bisected.

To prove that AD bisects BC in an isosceles triangle ABC, where AB = AC, follow these steps:

Step 1: Draw the given isosceles triangle ABC.
Step 2: Draw the altitude AD from the vertex A to the base BC.
Step 3: Label the intersection point of the altitude AD and the base BC as point E.
Step 4: Since the triangle ABC is an isosceles triangle with AB = AC, we can conclude that the angles opposite to AB and AC are equal. Therefore, angle B = angle C.
Step 5: As AD is an altitude, it is perpendicular to the base BC. Therefore, angle DAE = 90°.
Step 6: Since the angles of a triangle add up to 180°, we can calculate that angle BAC = 180° - angle B - angle C.
Step 7: Using the fact that angle B = angle C, we can simplify the equation to angle BAC = 180° - 2 * angle B.
Step 8: Similarly, we can calculate angle DAB as 180° - angle BAC.
Step 9: As mentioned earlier, angle B = angle C, and by substitution, we can simplify angle DAB to 180° - 2 * angle B.
Step 10: Since AD is an altitude, angle DAE = 90°. Therefore, angle BAE = 180° - angle DAE - angle DAB = 90° - (180° - 2 * angle B) = 2 * angle B - 90°.
Step 11: In triangle ABE, the angles add up to 180°. Therefore, angle A = 180° - angle BAE - angle BAE = 180° - (2 * angle B - 90°) - (2 * angle B - 90°) = 270° - 4 * angle B.
Step 12: In triangle ABC, the angles add up to 180°. Therefore, angle A + angle B + angle C = 270° - 4 * angle B + angle B + angle B = 180°.
Step 13: Simplifying the equation, we get 270° - 2 * angle B = 180°.
Step 14: Solving the equation, we find that 2 * angle B = 90°, which implies angle B = 45°.
Step 15: Since angle B = angle C, angle C is also 45°.
Step 16: In triangle ABC, angle A + angle B + angle C = 180°, substituting the known values, we get angle A + 45° + 45° = 180°, which implies angle A = 90°.
Step 17: Since angle A = 90°, triangle ABC is a right-angled triangle with AB = AC, making it an isosceles right-angled triangle.
Step 18: In a right-angled triangle, the altitude from the right angle to the hypotenuse bisects the hypotenuse. Therefore, AD bisects BC in the triangle ABC.
Step 19: Hence, AD bisects BC in the given isosceles triangle ABC with AB = AC.

To prove that AD bisects BC, we need to show that AD divides BC into two equal segments.

First, let's label the given information in the diagram. We have an isosceles triangle ABC, where AB = AC. AD is the altitude from vertex A to base BC.

To begin the proof, we will use the fact that in an isosceles triangle, the altitude from the vertex to the base divides the base into two equal segments.

Step 1: Draw the altitude AD from vertex A to base BC.

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