Question

In the given figure, AM = AN and angle1 = angle2.
Prove that:
(i) AngleAMC = AngleANC​

Answer 1

Answer:

Let us mark the points A and B on line l, C and D on line m and P and Q on line n.

Suppose the line n intersect line l at K and line m at L.

Since PQ is a straight line and ray KA stands on it, then

m∠AKP+m∠AKL=180

∘

(Angles in a linear pair)

⇒m∠a+m∠AKL=180

∘

⇒m∠a=180

∘

−m∠AKL ........(1)

Since PQ is a straight line and ray LD stands on it, then

m∠DLQ+m∠DLA=180

∘

(Angles in a linear pair)

⇒m∠b+m∠DLA=180

∘

⇒m∠b=180

∘

−m∠DLA ........(2)

Since, ∠a≅∠b, then m∠a=m∠b.

∴ from (1) and (2), we get

180

∘

−m∠AKL=180

∘

−m∠DLA

⇒m∠AKL=m∠DLA

⇒AKL≅∠DLA

It is known that, if a pair of alternate interior angles formed by a transversal of two lines is congruent, then the two lines are parallel.

∴AB∥CD or line l ∥ line m.

Answer 2

Step-by-step explanation:

So it is given that AM=AN and angle 1 = angle 2

First, we write this down

AM = AN

1 = 2 (Angles)

Then in the figure, we can see that the figures are 90 degrees so -

⛛ ANC = ⛛ AMC = 90 degree

Criteria = ASA

Because 2 angles and one side is given

SO BOTH THE TRIANGLES ARE EQUAL