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