In ∆ ABC ,AD is the median Show that AB²+AC²=2(AD²+BD²)
Answers
Answer:
construction :- draw a line AM perpendicular to BC.
we have to prove : AB² + AC² = 2(AD² + BD²)
proof :- case 1 :- when , it means AD is perpendicular on BC and both angles are right angle e.g., 90°
then, from ∆ADB,
according to Pythagoras theorem,
AB² = AD² + BD² ..... (1)
from ∆ADC ,
according to Pythagoras theorem,
AC² = AD² + DC² ...... (2)
AD is median.
so, BD = DC .......(3)
from equations (1) , (2) and (3),
AB² + AC² = AD² + AD² + BD² + BD²
AB² + AC² = 2(AD² + BD²) [hence proved ]
case 2 :- when
Let us consider that, ADB is an obtuse angle.
from ∆ABM,
from Pythagoras theorem,
AB² = AM² + BM²
AB² = AM² + (BD + DM)²
AB² = AM² + BD² + DM² + 2BD.DM ......(1)
from ∆ACM,
according to Pythagoras theorem,
AC² = AM² + CM²
AC² = AM² + (DC - DM)²
AC² = AM² + DC² + DM² - 2DC.DM ......(2)
from equations (1) and (2),
AB² + AC² = 2AM² + BD² + DC² + 2DM² + 2BD.DM - 2DC.DM
AB² + AC² = 2(AM² + DM²) + BD² + DC² + 2(BD.DM - DC.DM) ...........(3)
a/c to question, AD is median on BC.
so, BD = DC .....(4)
and from ADM,
according to Pythagoras theorem,
AD² = AM² + DM² ........(5)
putting equation (4) and equation (5) in equation (3),
AB² + AC² = 2AD² + 2BD² + 2(BD.DM - BD.DM)
AB² + AC² = 2(AD² + BD²) [hence proved].
Step-by-step explanation:
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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.!
