Prove that triangle awm is congruent to triangle cpm using a two column proof

ΔAWM ≅ ΔCPM proved

Step-by-step explanation:

See the given diagram with this question.

We have to prove ΔAWM ≅ ΔCPM

Now, (i) M is the mid point of AC, so AM =MC

(ii) Now, W is the mid point of AB, hence, AW = WB =

Again, P is the mid point of BC i.e

Now, it is given that BP = PM = MW = WB

Hence, BP = WB , hence, AW = PC

(iii) Again, PM = MW

Therefore, from the SSS rule, it is proved that ΔAWM ≅ ΔCPM. (Answer)

hope it you

step-by-step explanation:

sin(7pi/6 - pi/3)  

= sin(7pi/6 - 2pi/6)  

= sin(5pi/6)  

= sin(pi - pi/6)  

= sin(pi/6)  

= 1/2  

sin 7pi/6 - sin pi/3  

= sin (pi + pi/6) - sin pi/3  

= -sin pi/6 - sin pi/3  

= -1/2 - sqrt(3)/2  

= -(1+sqrt(3))2

d. corresponding side lengths in similar triangles are proportional in length.

step-by-step explanation: