Answer:
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From the given figure ,
RECA is a quadrilateral
RC divides it into two parts
From the triangles , ∆REC and ∆RAC
RE = RA (Given)
angle CRE = angle CRA (Given)
RC = RC (Common side)
Therefore, ∆REC is Congruent to ∆RAC
∆REC =~ ∆RAC by SAS Property
⇛CE = CA (Congruent parts in a congruent triangles)
Hence , Proved
<em>Additional comment:-</em>
SAS property:-
"The two sides and included angle of one triangle are equal to the two sides and included angle then the two triangles are Congruent and this property is called SAS Property (Side -Angle-Side)
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Answer:
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Step-by-step explanation:
Answer:
Finally, substitute AB = 6 and BC = 18 - 2AC in
AB + AC > BC 3rd Triangle Inequality condition
6 + AC > 18 - 2AC
-12 > -3AC
4 < AC
AC > 4
So putting all three results together
AC < 12, AC < 8, and AC > 4
Since AC is less than 8, it's automatically less than 12,
so we can dispense with AC < 12. So we have:
AC < 8 and AC > 4 which can be combined and written as
4 < AC < 8 [that is, AC is between 4 and 8]
Answer:
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Step-by-step explanation:
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