In triangle abc if ad is the median then show that ab^2+ac^2=2(ad^2+bd^2)
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Answer:
Step-by-step explanation:
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Answered by
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Answer:
AB² + AC² = 2(AD² + BD²).
Step-by-step explanation:
(i)
In ΔAED,
⇒ AD² = AE² + DE²
⇒ AE² = AD² - DE²
(ii)
In ΔAEB,
⇒ AB² = AE² + BE²
= AD² - DE² + BE²
= AD² - DE² + (BD + DE)² {BE = BD + DE}
= AD² - DE² + BD² + DE² + 2BD * DE - DE²
= AD² + BD² + 2BD * DE
(iii)
In ΔAEC,
⇒ AC² = AE² + EC²
= AD² - DE² + EC²
= AD² - DE² + (DC - DE)²
= AD² - DE² + DC² + DE² - 2DC * DE
= AD² + BD² - 2BD * DE {DC = BD}
On solving (ii) & (iii), we get
⇒ AB² + AC² = AD² + BD² + 2BD * DE + AD² + BD² - 2BD * DE
= AD² + BD² + AD² + BD²
= 2(AD² + BD)²
Hence proved.!
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