Question

please help me please yaar please ☆☆​

Answer 1

Answer:

See image

Step-by-step explanation:

Have Given :-

$\sf{Triangle \: ABC \: is \: an isosceles \: triangle.}$

$\sf \: AB = AC$

$\sf {AD \perp BC}$

To prove :-

$\sf{AD, median \: of \: the \: triangle}$

Proof :-

$\sf{In \: \: \triangle ADB \: \: and \: \: \triangle{ABC} \: -}$

$\sf{\angle{D} = \angle{D} \: \: (each 90°)}$

$\sf{AB = AC \: \: \: (Have \: Given)}$

$\sf{AD = AD \: \: \: (common)}$

So by RHS rule –

$\sf{\triangle{ADB} \: \cong \triangle{ADC}}$

$\therefore \: \: \sf{BD = CD}$

$\sf{ \therefore \: \: AD, \: is \: a \: median \: of \: \triangle ABC.}$

Hence proved

Answer 2

Answer:

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