The perpendicular AD on the base BC of a triangle ABC intersects BC at D so that BD=3CD . Prove that 2AB^2= 2AC^2+BC^2.
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Answers
Given :−
ABC is a triangle in which AD ⊥ BC and,
BD = 3CD
To prove :-
2AB² = 2AC² + BC²
Proof :-
we have,
BD = 3CD
therefore,
BC = (BD + CE) = 4CD
Now,
⤇ CD = ¼ BC .................................(i)
Then,
In ∆ ADB,
∠ ADB = 90°
By Pythagoras theorem,
AB² = AD² + BD² .........................(ii)
again,
In ∆ ADC,
∠ ADC = 90°
By Pythagoras theorem,
AC² = AD² + CD² .......................(iii)
Subtracting eq (iii) from (ii)
(AB²) - (AC)² = (AD² + BD²) - (AD² + CD²)
⤇ (AB² - AC²) = (BD² - CD²)
⤇ [(3CD)² - (CD)²]
⤇ 8CD² ⠀ ⠀ ⠀⠀ ⠀ ⠀ ⠀ ⠀ [ ∴ BD = 3CD ]
By using equation (i),
⤇ 8 × (1/16 BC²)
⤇ ½ BC²
∴ 2AB² - 2AC² = BC²
Hence, 2AB² = (2AC² + BC²)
Proved ....... :)
Answer:
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Step-by-step explanation:
Given that in ΔABC, we have
AD ⊥BC and BD = 3CD
In right angle triangles ADB and ADC, we have
AB^2 = AD^2 + BD^2 ...(i)
AC^2 = AD^2 + DC^2 ...(ii) [By Pythagoras theorem]
Subtracting equation (ii) from equation (i), we get
AB^2 - AC^2 = BD^2 - DC^2
= 9CD^2 - CD^2 [∴ BD = 3CD]
= 9CD^2 = 8(BC/4)^2 [Since, BC = DB + CD = 3CD + CD = 4CD]
Therefore, AB^2 - AC^2 = BC^2/2
⇒ 2(AB^2 - AC^2) = BC^2
⇒ 2AB^2 - 2AC^2 = BC^2
∴ 2AB^2 = 2AC^2 + BC^2.