If C is the incenter of ∆AMD, m∠A M C = 3 x + 6 and m∠ DMC= 8 x − 49. find ∠ M A D , find m∠ADM and find m∠ADC

3x+6=8x-49,

-6. -6

3x=8x-55

-8x. -8x

-5x=-55

x=11,

3(11)+6=39

8(11)-49=39

Step-by-step explanation:

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.