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If two sides of a triangle are unequal, ...

If two sides of a triangle are unequal, the longer side has greater angle opposite to it. GIVEN : A ` A B C` in which `A C > A Bdot` TO PROVE : `/_A B C >/_A C B`
CONSTRUCTION : Mark a point `D` on `A C` such that `A B=A Ddot` Joint `B Ddot`

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Given: A `/_\ABC` in which `AC>AB` (say)
To prove: `/_ABC>/_ACB`
Construction: Mark a point `D` on `AC` such that `AB=AD`. Join `BD`.
Proof: In `/_\ABD`
`AB=AD` (by construction)
`∠1=∠2` …(i) (angles opposite to equal sides are equal)
Now in `/_\BCD`
`/_2>/_DCB` (ext. angle is greater than one of the opposite interior angles)
`/_2>/_ACB` …(ii) `[since /_ACB=/_DCB]`
From (i) and (ii), we get
`/_1>/_ACB` …(iii)
But `/_1` is a part of `/_ABC`
`/_ABC>/_1` …..(iv)
Now from (iii) and (iv), we get
`/_ABC>/_ACB`
Hence proved.
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RD SHARMA-CONGRUENT TRIANGLE -All Questions
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