Read the proof. Given: AB ∥ DE Prove: △ACB ~ △DCE We are given AB ∥ DE. Because the lines are parallel and segment CB crosses bo
th lines, we can consider segment CB a transversal of the parallel lines. Angles CED and CBA are corresponding angles of transversal CB and are therefore congruent, so ∠CED ≅ ∠CBA. We can state ∠C ≅ ∠C using the reflexive property. Therefore, △ACB ~ △DCE by the AA similarity theorem. SSS similarity theorem. AAS similarity theorem. ASA similarity theorem.
We are given AB ∥ DE. Because the lines are parallel and segment CB crosses both lines, we can consider segment CB a transversal of the parallel lines. Angles CED and CBA are corresponding angles of transversal CB and are therefore congruent, so ∠CED ≅ ∠CBA.
Also ∠C ≅ ∠C using the reflexive property.
Therefore by AA similarity theorem , ΔACB ≈ ΔDCE
AA similarity theorem says that if in two triangles the two pairs of corresponding angles are congruent then the triangles are similar .
since the difference has to be 35 subtract 35 from 171 (equals 136) then divide it by 2, you get 68 which is your first number, then add 35 back and get 103 for the second number, I hope this helped
Ok, so there are 2 men per 7 women so we can just scale up so there are 4 men per 14 women and then 6 men per 21 women but there are 36 performers so we have to keep going till it equals 36 people so 8 men per 28 women = 36 people so 8 men!
