Question

In triangle ABC , ZB=90° and D is the mid-point of BC. Prove that
AC2 = AD2+3CD2​

Answer 1

Answer:

$\huge\purple {\mathfrak{Bonjour Mate!}}$

Given: In △ABC, ∠B = 90° and D is the mid-point of BC.

To Prove: AC2 = AD2 + 3CD2

Proof:

In △ABD,

AD2 = AB2 + BD2

AB2 = AD2 - BD2 .......(i)

In △ABC,

AC2 = AB2 + BC2

AB2 = AC2- BD2 ........(ii)

Equating (i) and (ii)

AD2 - BD2 = AC2 - BC2

AD2 - BD2 = AC2 - (BD + DC)2

AD2 - BD2 = AC2 - BD2- DC2- 2BDx DC

AD2 = AC2 - DC2 - 2DC2 (DC = BD)

AD2 = AC2 - 3DC2

Answer 2

Answer:

Step-by-step explanation:

D is the midpoint of BC= AD = CD.

Angle B is a right angled triangle.

Consider ΔABC

AC^2 = AB^2 + BC^2 [Pythagoras theorem]

⇒ AC2 = AB^2 + (2BD)^2  (BC=BD+DC,BD=DC,So we can replace BC by 2BD.

⇒ AC^2 = AB^2 + 4BD^2 ----------- (1)

Consider ΔABC

AD^2 = AB^2 + BD^2 [Pythagoras theorem] ----------- (2)

Subtracting equation (2) from (1), we get

⇒ AC^2 - AD^2 = 3BD^2

⇒ AC^2 - AD^2 = 3CD^2 [  BD = CD,So

Hope it helps you!