Question

The perpendicular from A on side BC of a triangle ABC of intersect BC at D Such that DB = 3CD then 2AC ( SQUARE ) + BC ( SQUARE ) is equal to
(a) AB (SQUARE )
(b) 3AB (SQUARE)
(c)2AB (SQUARE)
(d) 4AB (SQUARE)

Answer 1

Answer:

2AB(square) It helps you