Question

please explain similar triangles theorem

Answer 1

Draw 2 triangles ABC and DEF
Draw a line PQ in triangle DEF such tht PQ is parallel to EF

Now, AB = DP, BC = PQ, CA = EF

To Prove: Δ ABC ∼ Δ DEF

Draw PQ in triangle DEF so that, AB = DP and AC = DF

Proof:

ΔABC≅ΔDPQ and ΔABC≅ΔDPQ

Because corresponding sides of these two triangles are equal

ABDE=ACDF = ABDE=ACDF given

∠ A = ∠ D

Hence; ABDE=BCEF = ABDE=BCEF from SSS criterion

Hence;

ABDE = ACDF = BCEF = ABDE= ACDF = BCEF

Hence; Δ ABC ∼ Δ DEF proved

Answer 2

Your answer is in the pic.. attachment