Question

in fig ABC and AMP are two right triangl , right triangle angled at B and M respectively ​prove that triangle ABC is similar to triangle AMP

Answer 1

Answer:

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Answer 2

$\bold{\huge{\underline{\underline{\rm{ Given :}}}}}$

∆ ABC and ∆ AMP are right triangles at B and M respectively

$\bold{\huge{\underline{\underline{\rm{ To</strong><strong> </strong><strong> </strong><strong> </strong><strong> </strong><strong> </strong><strong> </strong><strong> </strong><strong> </strong><strong> </strong><strong> </strong><strong> </strong><strong> </strong><strong> </strong><strong>\</strong><strong>:</strong><strong> </strong><strong> </strong><strong>prove</strong><strong>:}}}}}$

$(i) in \triangle ABC and \triangle AMP$

$(ii) \frac{</p><p>CA }{</p><p> PA} = \frac{BC }{MP}$

$\bold{\huge{\underline{\underline{\rm{ proof :}}}}}$

In∆ ABC and ∆ AMP

$\angle </p><p>BAC = \angle MAP - common$

$\angle</p><p>ABC = \angle AMP - 90°</p><p> (given)$

$\therefore \triangle ABC \sim \triangle AMP$

....by AA

$(ii) \frac{CA }{PA} = \frac{BC }{MP}$.... ( corresponding parts of similar∆ triangle)