Please include what axioms and theorems were used as well.
2 answers:
Answer:
Hey!
In △ABD and△BAC,
AD = BC (it's given)
∠DAB = ∠CBA (it's given)
AB = BA (Common)
By SAS congruence rule,
∴△ABD≅△BAC
=> BD = AC (By C.P.C.T) (ii)
And, ∠ABD = ∠BAC (By C.P.C.T) (iii)
<em><u>Hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> </u></em><em><u>helps</u></em><em><u>,</u></em><em><u> </u></em><em><u>any</u></em><em><u> </u></em><em><u>confusions</u></em><em><u> </u></em><em><u>you</u></em><em><u> </u></em><em><u>may</u></em><em><u> </u></em><em><u>ask</u></em><em><u>.</u></em><em><u> </u></em>
<u>Part</u><u> </u><u>(</u><u>i</u><u>)</u>
1. ABCD is a quadrilateral in which AD=BC and ∠DAB=∠CBA (given)
2. AB = AB (reflexive property)
3. Triangles ABD and BAC are congruent (SAS)
<u>Part</u><u> </u><u>(</u><u>ii</u><u>)</u>
4. BD=AC (corresponding sides of congruent triangles are equal)
<u>Part</u><u> </u><u>(</u><u>iii</u><u>)</u>
5. ∠ABD = ∠BAC (corresponding angles of congruent triangles are equal)
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