Math, asked by sdmuddassir97, 11 months ago

- In ∆ABC,
seg AD I seg BC,
DB = 3CD.
Prove that 2AB^ = 2AC^+ BC^​

Answers

Answered by priyanshu3847
19

Step-by-step explanation:

In triangle abd

ab Sq. = ad Sq. + db Sq. .............1

in triangle and,

ac Sq. = ad Sq .+ cd Sq. ............2

subs. 1 and 2

ab sq -ac sq.=db square - cd sq. ..........3

=(3cd) sq.-cd sq.

= 9cd sq -cd sq.

=8× (1÷4 bc) sq.

=1/2 bc sq.

2 ab sq. - 2ac sq. = bc. sq.

2 ab sq. =2 ac sq. + bc sq.

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