Drag and drop the reasons into the boxes to correctly complete the proof.
Reason
Statement
Given
BD bisects LABC and AB BD
DBC ABD
Definition of angle bisector
2
3
AEB
DBC
Transitive Property of Congruence
LAEB ZABD
Alternate Interior Angles Theorem
ZABDZBAE
6
ZAEB
BAE
AB = EB
Converse of isosceles Triangle Theorem
7
8
AB= EB
Segment Congruence Postulate
10
=
Substitution Property of Equality
What is the answer!!!

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.

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step-by-step explanation: