Question

prove that the altitudes drawn on equal sides of an isosceles traingle are equal

Answer 1

Answer:

$\underline { Given }$

$ABC \:is \:a \: \blue { isosceles \: triangle }$

$AB = AC \: ---(1)$

$BD \:and \: CE \:are \: altitudes \: drawn \:to \\equal \:sides \: AC \:and \: AB \: respectively .$

$i.e\: \angle ADB = 90\degree , \: \angle ACE =90\degree \:---(2)$

$\underline { To \:prove: }$

$\red {BD = CF }$

$\underline { Proof}$

$In \: \triangle ABD \: and \triangle ACE ,$

$</p><p>[tex] \angle ADB = \angle ACE = 90\degree \: [from \: (2)]$

$\angle A = \angle A \: \pink { ( common \:angle )}$

$AB = AC \: [ From \:(1) ]$

$\therefore \triangle ABD \cong \triangle ACE$

$\blue {( AAS \: congruence \: Rule)}$

$BD = CE \: ( CPCT )$

$Hence, proved .$

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Answer 2

Answer:

prove that for an isosceles triangle altitudes drawn on the equal sides are equal