Given:
1 = 4
2 = 3

prove:
ab = cd

which of the following triangle congruence theorems would be used in this proof?

sss
asa
aas

 The answer is ASA congruence rule.

Step-by-step explanation:  We have the following rules for congruence of two triangles -

(i) SSS rule : If all the three sides of a triangle are equal to the corresponding three sides of the other triangle, then both of them are congruent by SSS rule.

(ii) ASA rule : If two angles and the side between them are equal to the corresponding two angles and the side between them of the other triangle, then the two triangles are congruent.

(iii) SAS rule : If two sides and the angle between them are equal to the corresponding two sides and the angle between them of the other triangle, then they are congruent.

Given that - In ΔABP and ΔCDQ (please the attached figure),

∠1 = ∠4,

∠2 = ∠3

and AB = CD.

Thus, ΔABP ≅ ΔCDQ by ASA rule.

Thus, the correct answer is ASA rule.

option a, c,d are correct.

step-by-step explanation:

from the given figure, it is given that z is equidistant from the sides of the triangle rst, then from triangle tzb and triangle szb, we have

tz=sz(given)

bz=zb(common)

therefore, by rhs rule,δtzb ≅δszb

by cpctc, sz≅tz

also, from δctz and δasz,

tz=sz(given)

∠tcz=∠saz(90°)

by rhs rule, δctz ≅ δasz, therefore by cpctc, ∠ctz≅∠asz

also,from δasz and δzsb,

zs=sz(common)

∠zbs=∠saz=90°

by rhs rule, δasz ≅δzsb, therefore, by cpctc, ∠asz≅∠zsb

hence, option a, c,d are correct.