Question

In ∆ABC,seg AD is a median then what is the ratio of A(∆ABD) and A(∆ADC).

Answer 1

Answer:

Step-by-step explanation:

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 , 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 ∠ABD≠ACD

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].