Question

In an obtuse triangle prove that AC²= AB²+BC²+2BC.BD​

Answer 1

CORRECT QUESTION :-

Triangle ABC is an obtuse triangle , obtuse angled at B.If AD is perpendicular to CB(produced). Prove that AC^2=AB^2+BC^2+2BC.BD.

ANSWER :-

in the right handed triangle ABD we have,

AD^2=AB^2-BD^2

in the right handed triangle ACD. WE have,

AD^2=AC^2-CD^2.

SO,

AC^2-CD^2=AB^2-BD^2.

=» AC^2=AB^2+CD^2 -BD^2.

=» =AB^2+(CD+BD)×(CD-BD)

=» =AB^2+BC ×(BC-CD) -(BC-BD)

=AB ^2 +BC ×( BC -2 BD)

=AB^2 -BC^2-2BC ×BD.

:)

Answer 2

Step-by-step explanation:

Given:- AD is perpendicular to BC and angle B is obtuse

To prove :- AC² = AB²+BC²+2BC*BD

Proof:- in ∆ADC,

AC²=AD²+DC² [pythagoras theorem ]

AC²=AD²+(DB+BC)²

AC²=(AD²+DB²)+BC²+2BC x BD

AC²=AB²+BC²+2BC x BD

Hence proved