Question

in in figure ABC is an isosceles triangle in which altitude be and CF are drawn to equal sides AC and AB respectively. show that these altitude are given​

Answer 1

$\sf \bf \huge {\boxed {\mathbb {QUESTION}}}$

$\tt In\: the\: given\: figure\: ABC\: is \:an\: isosceles \:triangle\\\tt in\: which\: altitudes\: BE\: and\: CF\: are\: drawn\: to\: equal\\\tt sides\: AC\: and\: AB\: respectively\:. Show \:that \:these\\\tt altitude\: are\: equal.$

$\sf \bf \huge {\boxed {\mathbb {ANSWER}}}$

$\sf \bf {\boxed {\mathbb {GIVEN}}}$

• $\sf \bf \triangle ABC\: is\: an\: isosceles\: triangle$
• $\sf \bf AB=AC$
• $\sf \bf BE\:and\:CF\:are\:altitudes$
• $\sf \bf \angle AEB={90}^{\circ}$
• $\sf \bf \angle AFC={90}^{\circ}$

$\sf \bf {\boxed {\mathbb {TO\:PROVE}}}$

• $\sf \bf BE=CF$

$\sf \bf {\boxed {\mathbb {SOLUTION}}}$

$\sf \bf In\: \triangle ABE\:and\: \triangle ACF$

$\sf \bf\implies \angle AEB=\angle AFC-(Given)$

$\sf \bf\implies \angle A=\angle A-(Common\:angle)$

$\sf \bf \implies AB=AC-(Given)$

$\sf \bf \therefore \triangle ABE\: \cong \: \triangle ACF-(AAS)$

$\sf \bf\implies BE=CF-(CPCT)$

$\tt {\underbrace {Hence\:Proved}}$

$\sf \bf \huge {\boxed {\mathbb {HOPE \:IT \:HELPS \:YOU}}}$

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$\sf \bf\huge {\boxed {\mathbb {EXTRA\:INFORMATION}}}$

$\tt {\overbrace{\underbrace {Congruent\:Triangle\:Postulates}}}$

$\sf \bf SSS=Side\:Side\:Side \:Postulate$

$\sf \bf SAS=Side\:Angle\:Side \:Postulate$

$\sf \bf ASA=Angle\:Side\:Angle\:Postulate$

$\sf \bf AAS=Angle\:Angle\:Side \:Postulate$

$\sf \bf RHS=Right\:angle\:Hypotenuse\:Side$

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