Question

∆ ABC is an equilateral ∆ . D is a point on line BC such that C is the midpoint of BD . let M be the midpoint of AD . If AB = 4 , then find BM^2.

the fig is attached above
please tell me the answer#with explanation
# spammers stay away​

Answer 1

Answer:

28

Step-by-step explanation:

joining M & C, as M & C are the midpoints so MC || AB ----(1)

Now, AC=BC=CD (Given) -----------(2)

by eq-(2), ∆ ACD Is isosceles (AC=CD) now, CM Is the median and we know that the median is also a perpendicular in isosceles triangle so angle CMD=90°

from eq-(1), angle BAM is also 90° ( corresponding angles)

in ∆ABM, AB^2+AM^2=BM^2 --------------(3)

now finding AM,

Constructing a perpendicular AK to BC.

AK^2+KD^2=AD^2

(AB^2 - BK^2) + (KC+CD)^2 = (2AM)^2

(4^2 - 2^2) + (2+4)^2 =4(AM)^2

(12+36)/4=AM^2

AM^2=12-------------(4)

Using eq--(3),

AB^2+AM^2=BM^2

4^2+12=BM^2 (Using eq-4)

BM^2=28

HOPE THIS WILL HELP

THANKS!

MARK BRILLIANT IF POSSIBLE.