Question

AD is a median of a triangle ABC and AM//BC. prove that: AC²=AD²+BC×DC+(BC/2)²​

Answer 1

Answer:

InΔAMC,

AC

2

=AM

2

+CM

2

[By Pythagoras Theorem]....(i)

InΔAMD,

AD

2

=AM

2

+DM

2

[By Pythagoras Theorem]....(ii)

∴AM

2

=AD

2

−DM

2

⇒AC

2

=[AD

2

−DM

2

]+CM

2

AC

2

=AD

2

−DM

2

+(DC+DM)

2

=AD

2

−DM

2

+DC

2

+2DC.DM+DM

2

=AD

2

+DC

2

+2DC.DM

=AD

2

+(

2

BC

)

2

+2

2

BC

DM

AC

2

=AD

2

+BC.DM+

4

BC

2

ii) From △ABM

AB

2

=AM

2

+BM

2

From △AMD,

AM

2

=AD

2

−DM

2

⇒AB

2

=[AD

2

−DM

2

]+BM

2

∴BM

2

=(BC−DC−DM)

2

(a−b−c)

2

=a

2

+b

2

+c

2

−2ab+2bc−2ac

AB

2

=AD

2

−DM

2

+[BC

2

+DC

2

−2(BC×DC)+2(DC×DM)−2(BC×DM)]

Put DC=

2

DC

∴AB

2

=AD

2

+BC

2

+

4

BC

2

−2

2

BC

2

+2(

2

BC

×DM)−2(BC×DM)

⇒AB

2

=AD

2

+

4

BC

2

−(BC×DM

WAS IT HELPFUL

Answer 2

Hope this answer will help you