In a triangle ABC, let D be the mid-point of BC. Prove that AB + AC > 2AD. (What property of quadrilateral is needed here?)

Open in App

Solution

Given: AD is the median of ΔABC.

To prove: AB + AC > 2AD

Construction: Produce AD to E such that AD = DE. Join CE.

Proof: In ΔABD and ΔCDE:

AD = DE (Construction)

∠ADB = ∠CDE (Vertically opposite angles)

BD = DC (AD is the median from A to BC)

∴ ΔABD ΔCDE (SAS congruence criterion)

⇒ AB = CE ... (1) (Corresponding parts of congruent triangles)

In ΔACE:

AC + CE > AE (Sum of any two sides of a triangle is greater than the third side)

⇒ AC + AB > AD + DE [Using (1)]

⇒ AC + AB > AD + AD (Construction)

⇒ AC + AB > 2AD

Suggest Corrections

Similar questions

Q. ABC is a triangle, D is the mid-point of BC. Prove that

\to A B + \to A C = 2 \to A D

Q. In triangle

A B C, D

is any point on side

B C

, then:

A B + B C + A C > 2 A D

Q. Question 6
ABC is a right triangle with AB = AC. If bisector of

∠ A

meets BC at D, then prove that BC = 2AD.

Q. If D is the midpoint of the side Bc of a triangle ABC, then prove that

\to A B + \to A C = \to 2 A D

Let ABCD be a quadrilateral and let P, Q, R, S be the mid-points of AB, BC, CD, DA respectively. Prove that PQRS is a parallelogram. (What extra result you need to prove this)?

Join BYJU'S Learning Program