The perpendicular AD on base BC of ∆ABC interestes BC at D, so that BD =3CD. Prove that 2AB^2=2AC^2+BC^2
Answers
Answer:
We have,
DB=3CD
∴ BC=BD+DC
⇒ BC=3CD+CD
⇒ BC=4CD
⇒ CD=
4
1
BC
∴ CD=
4
1
BC and BD=3CD=
4
3
BC.......(i)
Since △ABD is a right-angled triangle at D
∴ AB
2
=AD
2
+BD
2
......(ii)
Similarly, △ACD is a right-angled triangle at D.
∴ AC
2
=AD
2
+CD
2
.......(iii)
Subtracting equation (iii) from equation (ii) we get
AB
2
−AC
2
=BD
2
−CD
2
⇒ AB
2
−AC
2
=(
4
3
BC)
2
−(
4
1
BC)
2
[From (i) CD=
4
1
BC,BD=
4
3
BC]
⇒ AB
2
−AC
2
=
16
9
BC
2
−
16
1
BC
2
⇒ AB
2
−AC
2
=
2
1
BC
2
⇒ 2(AB
2
−AC
2
)=BC
2
⇒ 2AB
2
=2AC
2
+BC
2
[Hence proved]
Step-by-step explanation:
⌖GIVEN:-
⏣ A ∆ABC,in which AD perpendicular to BC and BD=3CD.
⌖TO PROVE:-
⏣ 2AB²=2AC²+BC²
⌖PROOF:-
We have :BD=3CD
∴ BC=(BD+CD)=4CD
➨ CD=1/4 BC ______( 1 )
In ΔADB ,∠ADB=90⁰
∴ AB²=AD²+CD² ______(2) [By pythagorus theorum]
In ΔADC ,∠ADC=90⁰
∴ AC²=AD²+CD² ______(3) [By pythagorus theorum ]
⎆On subtracting (3) from (2) ,we get,
➨(AB²-AC²)=(BD²-CD²)
=[(3CD)²-(CD)²] [⏣ BD=3CD]
=8CD²
=8×(1/16 BC²)
=1/2 BC² [⏣Using (1) ]
∴2AB²-2AC²=BC²
➨2AB²=(2AC²+BC²)
Hence, proved
