maths cbse class 10
Answers
Answer:
Step-by-step explanation:
Rohit2004rt · Helping Hand
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abhi178
Abhi178 The Sage
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].
Answer:
Step-by-step explanation:
AB∧2=AE∧2+BE∧2 EQ.1
AC∧2=AE∧2+EC∧2 EQ.2
ADDING EQ.1&2
AB∧2+AC∧2=2AE∧2+BC∧2
(but BC=2BD)
∴AB∧2+AC∧2=2(AE∧2+BD∧2)