Question
If $\triangle\text{ABC}$ and $\triangle\text{AMP}$ are two right triangles, right angled at B and M respectively such that $\angle\text{MAP}=\angle\text{BAC.}$ Prove that
  1. $\triangle\text{ABC}\sim\triangle\text{AMP}$
  2. $\frac{\text{CA}}{\text{PA}}=\frac{\text{BC}}{\text{MP}}$

Answer


  1. It is given that $\triangle\text{ABC}$ and $\triangle\text{AMP}$ are two right angle triangles.
Now, in $\triangle\text{ABC}$ and $\triangle\text{AMP}$, we have

$\angle\text{MAP}=\angle\text{BAC}$ (Given)

$\angle\text{AMP}=\angle\text{B}=90^\circ$

$\triangle\text{ABC}\sim\triangle\text{AMP}$ (AA Similarity)
  1. $\triangle\text{ABC}\sim\triangle\text{AMP}$
So, $\frac{\text{CA}}{\text{PA}}=\frac{\text{BC}}{\text{MP}}$ (Corresponding sides are proportional)

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