See the picture :D. Please help fast.
2 answers:
Answer:
In ΔABD and ΔAEC
BE=DC (given) …(i)
Subtracting DE from both side of (i), we get
BE–DE=DC–DE
BD=EC
AB=AC (given)
∠B=∠C (angle opposite to equal sides are equal)
Similarly, ΔABD≅ΔAEC
AD=AE (by c.p.c.t)
Hence proved
Answer:
Step-by-step explanation:
<u>Consider triangles ABE and ACD:</u>
- ∠BAC ≅ ∠CAB, common angle, congruent to itself
- ∠ABE ≅ ∠ACD, same arc DE intercepted
- AB = AC, given
<u>According to above congruence statements, we can conclude:</u>
- ΔABE ≅ ΔACD as per ASA postulate, since two angles and the included side are congruent
<u>Since the triangles are congruent, we have:</u>
- BE = CD as corresponding sides of congruent triangles
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