in equilateral triangle ABC BC perpenndicular to AD .such that D is point on BC then show that BC²=4(AC²-AD²)
Answers
Step-by-step explanation:
Here, triangle ABC is an equilateral triangle,
That is, AB = BC= CA
And, AD\perp BCAD⊥BC
Thus, by the property of equilateral triangle,
AD must be the bisector of side BC.
That is, BD = DC = \frac{1}{2} BCBD=DC=
2
1
BC
By the Pythagoras theorem,
AD^2 = AB^2 -BD^2AD
2
=AB
2
−BD
2
------(1)
And, AD^2 = AC^2 -CD^2AD
2
=AC
2
−CD
2
-------(2)
By adding equation (1) and (2),
We get,
2 AD^2 = AB^2 + AC^2 - (BD^2 + CD^2)2AD
2
=AB
2
+AC
2
−(BD
2
+CD
2
)
2 AD^2 =AC^2 + AC^2 - ((\frac{1}{2} BC)^2 + (\frac{1}{2} BC)^2)2AD
2
=AC
2
+AC
2
−((
2
1
BC)
2
+(
2
1
BC)
2
)
2 AD^2 = 2 AC^2 - (\frac{1}{4} BC^2 + \frac{1}{4} BC^2)2AD
2
=2AC
2
−(
4
1
BC
2
+
4
1
BC
2
)
2 AD^2 = 2 AC^2 - (\frac{1}{4} BC^2 + \frac{1}{4} BC^2)2AD
2
=2AC
2
−(
4
1
BC
2
+
4
1
BC
2
)
2 AD^2 = 2 AC^2 - \frac{1}{2} BC^22AD
2
=2AC
2
−
2
1
BC
2
2 AD^2 - 2 AC^2 = - \frac{1}{2} BC^22AD
2
−2AC
2
=−
2
1
BC
2
4 AD^2 - 4 AC^2 = - BC^24AD
2
−4AC
2
=−BC
2