Question

- In the given figure, sides AB and AC of AABC have been produced
to D and E respectively. If ZCBD = xº and ZBCE = yº such that
x > y, show that AB > AC.



plz dont give nonesence solution.​

Answer 1

$ANSWER</p><p></p><p>Since ∠ABC and ∠CBP form a linear pair.</p><p></p><p>∠ABC+∠CBP=180o</p><p></p><p>∠B+2∠1=180o           [BO is the bisector of ∠ CBP]</p><p></p><p>2∠1=180o−∠B</p><p></p><p>∠1=90o−21∠B       .....(1)</p><p></p><p></p><p>Again, ∠ACB and ∠QCB form a linear pair.</p><p></p><p>Therefore,</p><p></p><p>∠ACB+∠QCB=180o</p><p></p><p>∠C+2∠2=180o</p><p></p><p>2∠2=180o−∠C</p><p></p><p>∠2=90o−21∠C      ...(2)</p><p></p><p>In △BOC, we have</p><p></p><p>∠1+∠2+∠BOC=180o</p><p></p><p>90−21∠B+90o−21∠C+∠BOC=180o</p><p></p><p>180o−21(∠B+∠C)+∠BOC=180o</p><p></p><p>∠BOC=21(∠B+∠C)</p><p></p><p>∠BOC=21(180o−∠A)</p><p></p><p>∠BOC=90o−21∠A</p><p></p><p>$

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