prove in an isosceles triangle ABC,in which AB=AC and AD is the altitude,AD is also: i-median ii-bisector<A.
ADB is congruent to ADC --- side, side, side
the end really
To prove that in an isosceles triangle ABC, where AB = AC and AD is the altitude, AD is also:
i) the median, and
ii) the angle bisector of angle A
Let's begin the proof:
i) To prove that AD is the median of triangle ABC, we need to show that it divides the base BC into two equal parts.
Proof:
In triangle ABC, we know that AB = AC (as it is an isosceles triangle).
Let's assume that AD is not the median and it divides BC into two unequal parts, say BD and DC (where D is the point of intersection of AD and BC).
Case 1: Suppose BD < DC
Since AD is the altitude, it is perpendicular to BC. Hence, angle BAD = angle CAD (as they are both right angles).
Now, consider triangles ABD and ACD.
AD = AD (common side)
AB = AC (given)
Angle BAD = Angle CAD (as discussed above)
By the Side-Angle-Side (SAS) congruence criterion, triangles ABD and ACD are congruent.
Therefore, BD = DC (by corresponding parts of congruent triangles), which contradicts our earlier assumption (BD < DC).
Case 2: Suppose BD > DC
We can apply a similar argument as in Case 1 to show that BD = DC (contradiction).
Since both cases lead to contradiction, our assumption that AD is not the median is false. Hence, AD must be the median of triangle ABC.
ii) To prove that AD is the angle bisector of angle A, we need to show that it divides angle A into two congruent angles.
Proof:
In triangle ABC, let's consider angle A.
Since AD is the altitude, we know that angle BAD = angle CAD (as both are right angles).
Now consider triangles ABD and ACD.
AD = AD (common side)
AB = AC (given)
Angle BAD = Angle CAD (as discussed above)
Using the Side-Angle-Side (SAS) congruence criterion, triangles ABD and ACD are congruent.
Therefore, angle ADB = angle ADC (by corresponding parts of congruent triangles).
Hence, AD bisects angle A, proving that AD is the angle bisector of angle A.
This completes the proof that in an isosceles triangle ABC, where AB = AC and AD is the altitude, AD is also:
i) the median, and
ii) the angle bisector of angle A.