Question

In right triangle A EFG m2 E= 25°. In right triangle A HJK, MZ H=25º. Which
similarity postulate or theorem proves that A EFG and A HJK are similar?

Answer 1

Answer:

Therefore the triangles are congruent by Angle Angle similarity postulate.

ΔEFG  ~ ΔHJK ...{ By Angle-Angle similarity postulate}

Step-by-step explanation:

Given:

In right triangle ΔEFG

m∠E= 25°.

In right triangle ΔHJK,

m∠H=25º

To Prove:

ΔEFG  ~ ΔHJK

Proof:

In right triangle ΔEFG  and ΔHJK

m∠ E ≅ m∠ H .......{measure of each angle is 25° given}

m∠ F ≅ m∠ J .........{Both triangle is Right angle Triangle therefore measure angle is 90° each}

∴ ΔEFG  ~ ΔHJK ...{ By Angle-Angle similarity postulate}

Therefore the triangles are congruent by Angle Angle similarity postulate.

ΔEFG  ~ ΔHJK ...{ By Angle-Angle similarity postulate}

Answer 2

Answer:

Answer:

Therefore the triangles are congruent by Angle Angle similarity postulate.

ΔEFG  ~ ΔHJK ...{ By Angle-Angle similarity postulate}

Step-by-step explanation:

Given:

In right triangle ΔEFG

m∠E= 25°.

In right triangle ΔHJK,

m∠H=25º

To Prove:

ΔEFG  ~ ΔHJK

Proof:

In right triangle ΔEFG  and ΔHJK

m∠ E ≅ m∠ H .......{measure of each angle is 25° given}

m∠ F ≅ m∠ J .........{Both triangle is Right angle Triangle therefore measure angle is 90° each}

∴ ΔEFG  ~ ΔHJK ...{ By Angle-Angle similarity postulate}

Therefore the triangles are congruent by Angle Angle similarity postulate.

ΔEFG  ~ ΔHJK ...{ By Angle-Angle similarity postulate}

Step-by-step explanation: