Question

plz answer this question ​

Answer 1

Answer:

plz hit thank if it helps

Step-by-step explanation:

1st part .......... in figure CE parallel to AB and AC is a transversal so

angle BAC = angle ACE (alternate so angle ACE = 40°

2nd part .... now in triangle angle A+ angle B + angle C = 180

40° + 70° + angle C = 180

110 + angle C = 180

Angle C = 70 °

angle ACB = 70

now

Angle ACB + ACE + ECD = 180 ( angle on straight line )

70 + 40 + ECD = 180

ECD = 70

3rd part ......... now angle ACD =

angle ACE + CED

so 70 + 70 = 110

Answer 2

Step-by-step explanation:

To Prove ABC + ACD = 2A C

AL in the angle bisector of A, so let BAL = LAC = x

Then, using exterior angle property

∠ALC = x + alpha ..(1)

Similarly

∠ACD= 2x + ..(2)

Hence from (1) and (2), we get

ABC + ACD = 2ALC

(i.e.)

( alpha ) + (2x + alpha ) = 2(x + alpha )

HENCE PROVED