Question

17.Prove the angles opposite to equal sides of a triangle are equal.
Pls help me with this

Answer 1

$\huge\sf\blue{Given}$

✭ Opposite sides of a triangle are equal

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$\huge\sf\gray{To \:Prove}$

◈ The angles opposite to equal sides are equal

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$\huge\sf\purple{Steps}$

$\sf \underline{\underline{\sf Concept}}$

Let's consider a ∆ABC, where AB = AC and we got to prove that ∠C = ∠B and so we can simply prove it by proving that two triangles are equal

$\sf \underline{\underline{\sf Construction}}$

»» Draw AD ⊥ CB

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So in ∆ACD & ∆ABD

➝ $\sf AC = AB$ «« Given »»

➝ $\sf \angle CDA = \angle BDA$ «« Construction (90°) »»

➝ $\sf AD = AD$ «« Common »»

$\sf \therefore \triangle ACD \cong \triangle ABD$

So now we may conclude that,

➳ $\sf \angle C = \angle B$ «« CPCT »»

$\sf Hence \ Proved!!$

$\sf\star\: Diagram \:\star$

$\setlength{\unitlength}{30} \begin{picture}(6,6)\thicklines\put(2,1){\line(1,0){5}}\put(2,1){\line(1,1){2.5}}\put(7,1){\line(-1,1){2.5}}\put(4.5,1){\line(0,1){2.5}}\put(2,0.5){ \tt C }\put(7,0.5){ \tt B }\put(4.5,3.7){ \tt A }\put(4.5,0.5){ \tt D }\end{picture}$

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