Suppose ABCD is a rhombus such that the angle bisector of ∠ABD meets
AD at point K. Prove that m∠AKB = 3m∠ABK. Help meee!! What do I put in the box shown below??

you can not drow on this

step-by-step explanation:

step by step explanation: when u removing the parentheses and combined them that would be : x+3y-x-y=2y

55 because 55 + 55 = 110 good luck babe

part a: triangles gcb and epb

part b. sides cg and rp are parallel by definition of parallelogram.

angles cgb and e are congruent by being alternate interior angles of parallel lines cut by a transversal. angles gbc and ebp are congruent because they are vertical angles. by aa similarity, the triangles are similar.

part c. since the triangles are similar, the ratios of the lengths of corresponding sides are equal.

calculation of be:

gb/be = cb/bp

(400 ft)/be = 300/200

(400 ft)/be = 3/2

3be = 2 * 400 ft

3be = 800 ft

be = 800/3 ft

calculation of pe:

gc/pe = cb/bp

(350 ft)/pe = 300/200

(350 ft)/pe = 3/2

3pe = 2 * 350 ft

3pe = 700 ft

pe = 700/3 ft