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State true or false.

In an equilateral triangle $A B C, A D$ is drawn perpendicular to $B C$ , meeting $B C$ in $D$ . Then $A D 2 = 3 B D 2$ .

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True

False

answer is A.

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Detailed Solution

Given that, in an equilateral triangle

A B C, A D

is drawn perpendicular to

B C

Apply the Pythagoras theorem in the triangle

A B D

(b a s e) 2 + (p e r p e n d i c u l a r) 2 = (h y p o t e n u s e) 2

⇒ A B 2 = A D 2 + B D 2

All three sides of a triangle are equal since the triangle is equilateral. So,

A B = B C

.
Then,

B C 2 = A D 2 + B D 2

. ……(1)
Solve equation (1).
Since

A D

is perpendicular to the line

B C

and perpendicular in an equilateral triangle is the median.
So,

B C = B D + D C

.
Therefore,

⇒ B C = B D + B D ⇒ B C = 2 B D ⇒ B C 2 = 2 B D 2

……(2)
Replacing the value from equation (2) to equation (1).

⇒ 2 B D 2 = A D 2 + B D 2 ⇒ 4 B D 2 = A D 2 + B D 2 ⇒ 4 B D 2 − B D 2 = A D 2 ⇒ 3 B D 2 = A D 2

So,

A D 2 = 3 B D 2

.
Hence, the statement is true.

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