Question

If ∆ABC and ∆AMP are two right triangles, right angled at B and M. Prove that ∆ABC =∆AMP

Answer 1

$\huge{\boxed{Question:-}}$

If ∆ABC and ∆AMP are two right triangles, right angled at B and M. Prove that ∆ABC=∆AMP.

$\huge{\boxed{Solution:-}}$

In ∆ABC:-

/_ ABC = 90°

/_ BAC = /_ ACB

Let /_ BAC = /_ ACB = x

By angle sum property of a triangle:-

/_ ABC + /_ BAC + /_ ACB = 180°

${90°} + {x} + {x} = {180°}$

${2x} = {180°} - {90°}$

${2x} = {90°}$

${x} =\frac{90°}{2}$

${x} = {45°}$

/_ BAC = /_ ACB = 45°

In ∆AMP:-

/_ AMP = 90°

/_ MAP = /_ APM

Let /_ MAP = /_ APM = x

By angle sum property of a triangle:-

/_ AMP + /_ MAP + /_ APM = 180°

${90°} + {x} + {x} = {180°}$

${2x} = {180°} - {90°}$

${2x} = {90°}$

${x} =\frac{90°}{2}$

${x} = {45°}$

/_ MAP = /_ APM = 45°

$\huge{\boxed{\mathfrak{\red{Answer:-}}}}$

∆ABC ≈ ∆AMP by AAA rule of congruency.