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Similarity of Mathematical Forms in Grade Two of Junior High School
(1) According to the proportional relationship of similar triangles, we get:

Height of the tree =1.6 * 4.5 = 7.2m.

(2) Make the parallel lines of AG intersect with the extension lines of CB at g..

So: OE is the center line of the triangle AGC, GE=EC, let BE=X, then: EC = BC-be = b-x, BG = ge-be = (b-x)-x = b-2x.

Triangle ABG is similar to triangle February.

Blood sugar /BE=AB/BF

(b-2X)/X = air conditioning

X=bc/(a+2c)