Chapter 7 theorem 7.6
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Gaurav Seth 5 years ago
“If two sides of a triangle are unequal then the longer side has greater angle opposite to it.”
Given: A ΔABC in which AC > AB (say)
To prove: ∠ABC > ∠ACB
Construction: Mark a point D on AC such that AB = AD.
Join BD.
Proof: In ΔABD
AB = AD (by construction)
⇒ ∠1 = ∠2 …(i) (angles opposite to equal sides are equal)
Now in ΔBCD
∠2 > ∠DCB (ext. angle is greater than one of the opposite interior angles)
⇒ ∠2 > ∠ACB …(ii) [∵ ∠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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