Question

ABC is an equilateral triangle and D is any point in AC. Prove that BD>AC.

Answer 1

BD > AD in a triangle ABC

Step-by-step explanation:

Given Data

ABC is an equilateral triangle and D is any point in AC

To prove - BD > AD

The triangle in which all the three sides are called as Equilateral triangle.

Let us consider an equilateral triangle ABC.

D is the any point in AC.

Now inside the triangle ABC, there were two triangles namely triangle ADB or ABD and triangle BDC or DBC

where ∠BAD = 60°

from the figure, it is clear that ∠ABD is is lesser than ∠ABC

which means  ∠ABC is greater than 60° and ∠ABD <60°

Now in this triangle ABC, opposite side larger angles are greater than opposite side smaller angles, with BD opposite to ∠BAD , AD opposite to ∠ABD

∠ABD < ∠BAD = 60°

BD > AD

Therefore It is prove that BD > AC in an equilateral triangle ABC with D is any point in AC.

Refer the figure for better understanding.

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Answer 2

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CBSE

Mathematics

Grade 10

Triangles

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ABC is an equilateral triangle and D is any point in AC. Prove that BD > AD.

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Hint: We will first start by using the fact that ΔABC is an equilateral triangle. Therefore, the angles of ΔABC are 60∘ each. Then we will prove that ∠BAD<∠ABD and will use the property of triangle that side opposite larger angle is greater than the side opposite smaller angle.

Complete step-by-step answer:

Now, we have been given that ABC is an equilateral triangle and D is any point in AC and we have to prove that BD > AD.

Now, we know that in an equilateral triangle each angle is 60∘. Therefore, we have ∠BAC=60∘ and ∠ABC=60∘.

Now, we have from the figure that,

∠ABC=∠ABD+∠DBC∠ABD+∠DBC=∠ABC

Now, we will substitute ∠ABC=60∘.

∠ABD+∠DBC=60∘

Or we can say that,

∠ABD<60∘

Now, we will substitute 60∘=∠BAD. So, we have,

∠ABD<∠BAD

Now, we know that the side opposite to larger angle is greater than the side opposite to smaller angle. So, we have,

AD<BD

Hence Proved.