Question

angle B of triangle ABC is an acute angle and AD is perpendicular on BC. prove that AC^2 = AB^2 + BC^2 - (2 BC)(BD)

Answer 1

hello,
inΔADC-
AC²=AD²+DC²(pythagoras theorem)...................1
in ΔADB-
AB²=AD²+BD²(samee reason).............................2
AC²=AD²+(BC-BD)²  (BD+DC=BC) from 1
AC²=AD²+BC²+BD²-2·BC·BD [(a-b)²=a²+b²-2ab]
AC²=(AD²+BD²)+BC²-2·BC·BD
from 2
AC²=AB²+BC²-2.BC.BD
hence proved

Answer 2

Given,

Angle B = Acute angle

AD is perpendicular upon BC.

To find,

Proving, (AC)² = (AB)²+(BC)²- 2×(BC×BD)

Solution,

We have to take references of the Pythagoras theorem, in this case.

According, to the data mentioned in the question, we can say that ∆ABC and ∆ADC are right angled triangles.

So,

(AB)² = (AD)²+(BD)²

And,

(AC)² = (AD)²+(CD)²

or, (AC)² = (AD)² + (BC-BD)² [Because, CD+BD = BC)

or, (AC)² = (AD)² + (BC)²+(BD)²-2×(BC×BD)

or, (AC)² = (AD)²+(BD)²+(BC)²-2×(BC×BD)

or, (AC)² = (AB)²+(BC)²-2×(BC×BD)  [proved]

(Diagram is attached below, for better understanding.)

Hence, the given statement is proved by the above mentioned discussion.