given ABC is a triangle. Ad is the altitude to BC. Ad biscects angle BAC.
Prove: Ad is the median to BC
Plz tell me by using section formula
Consider triangles ABd and ACd for congruence:
Ad is altitude, so
∠ABd = ∠ACd = 90°
Ad bisects angle BAC, so
∠CAd = ∠BAd
Ad = Ad (common side)
So triangles ABd and ACd are congruent, ASA (angle-side-angle).
To prove that AD is the median of triangle ABC, we need to show that AD divides BC into two equal segments.
Let's proceed with the proof:
Step 1: Given that AD is the altitude to BC, we know that it forms a right angle with BC at point D.
Step 2: Since AD bisects angle BAC, this means that angle BAD is congruent to angle CAD. This follows from the definition of angle bisector.
Step 3: Let's consider triangle ADB and triangle ADC. We can prove that they are congruent triangles using the Angle-Side-Angle (ASA) congruence criterion.
- Angle ADB is equal to angle ADC since they are vertical angles.
- Angle BAD is congruent to angle CAD (from step 2).
- Side AD is common to both triangles.
Therefore, triangle ADB is congruent to triangle ADC by ASA criterion.
Step 4: Since triangle ADB is congruent to triangle ADC, their corresponding sides are also congruent. So, AB is congruent to AC.
Step 5: Now, consider the segment BC. Since AD is the altitude, we know that angle BDA and angle CDA are both right angles.
Step 6: By the definition of a median, it connects the midpoint of one side to the opposite vertex. Let's call the midpoint of BC as M.
Step 7: Since AD is the altitude, it meets BC at a right angle at point D. This implies that DM is equal to MC since M is the midpoint of BC.
Step 8: From step 4, we know that AB is congruent to AC. Thus, triangle ABC is an isosceles triangle.
Step 9: In an isosceles triangle, the altitude drawn from the vertex bisects the base. Therefore, AD bisects BC, and we can conclude that DM is equal to MC.
Hence, we have proven that AD is the median of triangle ABC as it divides BC into two equal segments.
To prove that AD is the median of triangle ABC, we need to show that it divides the side BC into two equal segments.
In order to do this, we can use the properties of a triangle and angle bisectors. Let's break down the proof into two parts:
Part 1: Angle Bisector Property
Since AD is the angle bisector of angle BAC, it divides angle BAC into two equal angles (angle BAD and angle CAD). This property can be stated as:
1. Angle BAD = Angle CAD
Part 2: The Side-Splitter Theorem
Now, let's consider triangle ABC. Since AD is an altitude to side BC, it is perpendicular to BC. This property can be stated as:
2. Angle ADB = Angle ADC = 90 degrees
Now, we can use the angle bisector property from Part 1 to show that triangles ABD and ACD are similar. This is because angle BAD = angle CAD and angle ADB = angle ADC = 90 degrees.
Using the Angle-Angle (AA) similarity theorem, we can say that triangles ABD and ACD are similar. This similarity leads to the side-splitter theorem, which states that the lengths of two sides of similar triangles are proportional.
Since AD is an altitude, it divides BC into two segments, BD and CD. By the side-splitter theorem, we can conclude:
3. BD/DC = AB/AC
Since triangle ABC is an isosceles triangle (AD is an altitude and AD bisects angle BAC), we know that AB = AC. Therefore, we can substitute AB with AC in equation 3:
4. BD/DC = AC/AC
Simplifying equation 4:
5. BD/DC = 1
This tells us that BD is equal in length to DC, which means that AD divides BC into two equal segments. Hence, AD is the median of triangle ABC.
Therefore, we have proved that AD is the median to BC in triangle ABC using the angle bisector property and the side-splitter theorem.