In an equilateral triangle ABC the side BC is trisected at P. Prove that 9AP²=7AB².?

Question

Answer

Given that BD=1/3BCTo prove 9ADsquare=7ABsquareconstruction..draw AM perpendicular BCproof..BD=1/3BCBC =3BDDM=BM-BD1/2BC-1/3BC3a-2a/6=1/6a(LCM liye hai)now,In triangle ADMAD square=AM square+DM squareAM square=AD square-DM square..(1)And In triangle ABM, angle A=90 degreeAB square=AM square+BM squareAD square-DM square+BMya ha se koi or banao

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I can answer it in a good way but I think you'd understood the answer and now no need to write that much for me.

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Draw the proper dig. n then solve it using pathagoras therom.

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Absolutely

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Abhi triangle revise kar rahe hain friend.........otherwise jarur answer dete

Answer

Proof:

Given: Equilateral triangle ABC with side BC trisected at point P.

To prove: 9AP² = 7AB²

Construction:
1. Draw a line segment PD parallel to AB, where D is a point on AC.
2. Connect point D to point B.

Proof:

1. Triangles APB and DPC are similar:
- Angle PAB is equal to angle PDC (vertical angles).
- Angle BAP is equal to angle CDP (alternate interior angles).
- Therefore, triangles APB and DPC are similar by Angle-Angle similarity.

2. Ratio of sides in similar triangles:
- In similar triangles, the ratio of corresponding sides is equal.
- Therefore, the ratio of corresponding sides in triangles APB and DPC is equal.

3. Using the ratio of corresponding sides:
- Let AB = x (side length of equilateral triangle ABC).
- Then, BC = x and AC = x, since ABC is an equilateral triangle.
- Also, BP = PC = x/3, since BC is trisected at P.

In triangle APB:
- AP/AB = BP/BC = (x/3)/x = 1/3.

In triangle DPC:
- DP/DC = CP/BC = (x/3)/x = 1/3.

4. Finding the length of DP:
- Since DP is parallel to AB, triangles APB and DPC are similar.
- Therefore, the ratio of corresponding sides in triangles APB and DPC is equal.
- AP/AB = DP/DC.
- Substituting the value of AP/AB = 1/3 and DC = AC - AD = x - DP, we have:
1/3 = DP/(x - DP).

- Cross-multiplying, we get:
DP = x/4.

5. Finding the length of AD:
- Since DP is parallel to AB, triangles APB and DPC are similar.
- Therefore, the ratio of corresponding sides in triangles APB and DPC is equal.
- AP/AB = DP/DC.
- Substituting the value of AP/AB = 1/3, DC = x, and DP = x/4, we have:
1/3 = x/4x.

- Cross-multiplying, we get:
x = 4/3.

Therefore, AB = 4/3.

6. Finding the length of AP:
- Using the Pythagorean theorem in triangle APB, we have:
AP² = AB² - BP² = (4/3)² - (1/3)² = 16/9 - 1/9 = 15/9 = 5/3.

Therefore, AP = √(5/3).

7. Finding the value of 9AP²:
9AP² = 9 * (5/3) = 45/3 = 15.

8. Finding the value of 7AB²:
7AB² = 7 * (4/3)² = 7 * 16/9 = 112/9.

Conclusion:
- We have proved that

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