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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Answer:
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Step-by-step explanation:
(6, -7) and (8,5)
m=(y2-y1)/(x2-x1)
m=(5 + 7)/(8 - 6)
m= 12/2
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y + 7 = 6(x - 6)
y + 7 = 6x - 36
y = 6x - 43
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Answer:
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Step-by-step explanation:
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