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Answer :
Answer:
To prove that the altitudes to the legs of an isosceles triangle are congruent, we can use the ASA (Angle-Side-Angle) rule for proving triangles congruent. Here's the proof:
Let's consider an isosceles triangle ABC where AB = AC, meaning that the two sides adjacent to the base are congruent. We want to prove that the altitudes drawn from points B and C to the opposite sides are congruent.
Proof:
1. Draw the altitude from vertex B to side AC, and let's call the point where it intersects side AC as D.
2. Draw the altitude from vertex C to side AB, and let's call the point where it intersects side AB as E.
Now, we have two right triangles, ABD and ACE, where:
- AD and AE are the altitudes from B and C, respectively.
- AB = AC (given that it's an isosceles triangle).
- BD = CE (both are altitudes, and they are equal in length).
- Angle ADB = Angle AEC = 90 degrees (as they are right angles).
By the ASA rule:
- Angle ADB = Angle AEC (by definition).
- AB = AC (given).
- BD = CE (given).
Therefore, by the ASA rule, triangles ABD and ACE are congruent.
Since the triangles are congruent, their corresponding parts are also congruent. In particular, AD (the altitude from B) and AE (the altitude from C) are congruent.
So, we've proven that the altitudes to the legs of an isosceles triangle are congruent using the ASA rule for triangle congruence.
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Step-by-step explanation:
To prove that the altitudes to the legs of an isosceles triangle are congruent, you can use the rule ASA (Angle-Side-Angle). Here's the proof:
Given: We have an isosceles triangle ABC, with AB = AC.
To prove: The altitudes from B and C to the legs are congruent.
Proof:
1. Draw an altitude from vertex B to side AC, and label the point of intersection as D.
2. Draw an altitude from vertex C to side AB, and label the point of intersection as E.
We need to prove that segment DE is congruent to segment DF, which will establish that the altitudes from B and C are congruent.
3. In triangle ABD and triangle ACE:
- Angle ABD = Angle ACE (both are right angles as they are altitudes).
- AB = AC (given that it's an isosceles triangle).
- AD = AE (both are altitudes, so they are perpendicular to the respective bases).
4. By the ASA rule, we can conclude that triangle ABD is congruent to triangle ACE.
5. Since these triangles are congruent, their corresponding parts are congruent.
- BD = CE (corresponding parts of congruent triangles).
6. Now, notice that DE is equal to the sum of BD and CE (DE = BD + CE).
7. Substituting the values, DE = BD + CE = CE + CE = 2CE.
8. Thus, DE is twice the length of CE (DE = 2CE).
9. Similarly, DF is twice the length of CF (DF = 2CF).
10. Since DE = 2CE and DF = 2CF, we can conclude that DE = DF, meaning that the altitudes from B and C are congruent.
So, by proving that DE is congruent to DF, you've established that the altitudes to the legs of an isosceles triangle are congruent.
