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In triangle ABC, AD is perpendicular to BC and BD is equal to one third of CD prove that 2 AC square is equal to 2AB square BC square?
Most Upvoted Answer
In triangle ABC, AD is perpendicular to BC and BD is equal to one thir...
In right ADB, AB²= AD² + BD² [By Pythagoras theorem] – (1)
In right ADC, AC²= AD² + CD² [By Pythagoras theorem] – (2)
Subtract (1) from (2),
we get AC² – AB2 = AD2 + CD2 – AD2 – BD2
AC2 – AB2 = CD2 – BD2 --- (3)
Given BD =1/3 CD
From the figure, BD + DC = BC
⇒ (1/3)CD + CD = BC
⇒ (4/3)CD = BC
∴ CD = 3BC/4
⇒ BD = BC/4
Xf Equation (3) becomes, AC2 – AB2 = (3BC/4)2 – (BC/4)2 = (8/16)BC2
⇒ AC2 – AB2 = (1/2) BC2
⇒ 2AC2 – 2AB2 = BC2
⇒ 2AC2 = 2AB2 + BC2
Community Answer
In triangle ABC, AD is perpendicular to BC and BD is equal to one thir...
Proof:
1. Given:
- In triangle ABC, AD is perpendicular to BC.
- BD = 1/3 * CD
2. To prove:
- 2AC^2 = 2AB^2 + BC^2
3. Proof:
- Let's denote the lengths of the sides as follows:
- AC = x
- AB = y
- BC = z
4. By Pythagorean theorem:
- In triangle ABD, AD^2 + BD^2 = AB^2
- x^2 + (1/3 * z)^2 = y^2
- x^2 + (1/9)z^2 = y^2
5. By Pythagorean theorem:
- In triangle ACD, AD^2 + CD^2 = AC^2
- x^2 + z^2 = AC^2
6. Substitute the value of AC^2:
- x^2 + z^2 = 2AB^2 + BC^2
- 2AB^2 + BC^2 = 2AC^2
7. Hence, proved that:
- 2AC^2 = 2AB^2 + BC^2
8. Conclusion:
- The given relationship holds true in triangle ABC with the given conditions.
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In triangle ABC, AD is perpendicular to BC and BD is equal to one third of CD prove that 2 AC square is equal to 2AB square BC square?
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