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zubka84 [21]
3 years ago
11

In the fig. AD = DC and AB = BC. Prove that triangle ABD = triangle CDB​

Mathematics
1 answer:
nexus9112 [7]3 years ago
7 0

<u>Statement                        |   Reason</u>

AD = DC                           |   given

AB = CB                           |    given

BD = DB                           |   reflexive property

ABD = CDB                      |    side-side-side congruency theorem

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4 0
3 years ago
Read 2 more answers
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6 0
3 years ago
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Vsevolod [243]

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9.\\7\dfrac{1}{6}-2\dfrac{5}{6}=6\dfrac{6+1}{6}-2\dfrac{5}{6}=6\dfrac{7}{6}-2\dfrac{5}{6}=(6-2)+\dfrac{7-5}{6}=4\dfrac{2}{6}=\boxed{4\dfrac{1}{3}}\\\\10.\\9\dfrac{3}{12}-4\dfrac{7}{12}=8\dfrac{12+3}{12}-4\dfrac{7}{12}=8\dfrac{15}{12}-4\dfrac{7}{12}=(8-4)+\dfrac{15-7}{12}=4\dfrac{8}{12}\\\\=4\dfrac{8:4}{12:4}=\boxed{4\dfrac{2}{3}}

5 0
3 years ago
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