Question

Which theorem or postulate proves that △ABC and △DEF are similar?

Answer 1

For similarity measures of corresponding angles should be equal so we would use AA similarity postulate

Answer 2

Answer:

AA similarity postulate holds here.

Step-by-step explanation:

As in right angled ΔCAB and ΔFDE we are given two angles each and with the help of property that in a triangle the sum of all the angles of a triangle is equal to 180° w can find out the measure of third angle.

In ΔCAB we have:

∠A=90° and ∠B=37°.

The measure of third angle i.e. ∠C will be:

∠A+∠B+∠C=180°

90°+37°+∠C=180°

∠C=180°-127°

⇒  m∠C=53°

Similarly in ΔFDE we get:

m∠E=37° ( since we are given ∠F=53° and ∠D=90°)

Hence clearly all the corresponding  angles in both the triangles are equal.

Hence AAA similarity holds.

Hence, both the triangles are similar.

Hence, first option is true.

AA similarity postulate