Question

AD is a median of ΔABC. If BD = AD, prove that ∠A is a right angle in ΔABC.

Answer 1

in ABC, use Apolloneous’ theorem,
we know that,
For AD to be median, we have,
AB² + AC² = 2(AD² + BD²)
as it is given that AD = BD
AB² + AC² = 2(BD² + BD²)
AB² + AC² = 4BD²
since, AD is the median to line BC , we have,
BD = BC/2
now, substituting the value of BD in above equation,
AB² + AC² = 4 × {BC/2}²
AB² + AC² = 4 × BC²/4 = BC²
AB² + AC² = BC²
from converse of Pythagoras theorem,
∆BAC is right angled triangle where A is right angle.

Answer 2

In ∆ABC , AD is the median .

BD = DC -----( 1 )

BD = AD ---( 2 ) ---( given )

i ) In ADB ,

BD = AD [ from ( 2 ) ]

<BAD = <DBA = x [ angles opposite

to equal angles ]

ii )

CD = BD [ from ( 1 ) ]

CD = AD [ from ( 2 ) ]

In ∆ADC ,

AD = CD

<DAC = <ACD = y

[ angles opposite to equal sides ]

iii ) In ABC ,

<BAC+ <ABC + <ACB = 180°
( x + y ) + x + y = 180°

2( x + y ) = 180°

x + y = 90°

Therefore ,

<BAC = x + y = 90°

∆ABC is a right angled triangle.

I hope this helps you.

: )