Math, asked by Aksharasree8151, 1 year ago

In triangle ABC if AD is the median then show that AB2 + AC2 = 2(AD2 + BD2)

Answers

Answered by abhi178
698
see figure, here it is clearly shown that ABC is a triangle, in which AD is a median on BC.

construction :- draw a line AM perpendicular to BC.

we have to prove : AB² + AC² = 2(AD² + BD²)

proof :- case 1 :- when \angle{ADC}=\angle{ADB}, 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)

\because 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 \angle{ADB}\neq\angle{ADC}
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].
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Answered by rithudevarasetty
272

Answer:

Step-by-step explanation:

AD is median, So BD=DC.

AB2= AE2+BE2 (1)

AC2 = AE2+EC2 (2)

Adding both equation 1&2,

We get,

AB2+AC2 = 2AE2+BE2+CE2

= 2(AD2-ED2)+(BD-ED)2+(DC+ED)2

= 2AD2-2ED2+BD2+ED2-2BD.ED+DC2+ED2+2CD.ED

= 2AD2+BD2+CD2

= 2(AD2+ BD2)

Hence proved.......

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