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In triangle ABC if AP perpendicular BC and AC square=BC square-AB square then prove that PA square=PB square*CP square

Posted by Kkk M 5 years, 9 months ago

CBSE > Class 10 > Mathematics

1 answers

Sia ? 5 years, 9 months ago

According to question it is given that in ∆ ABC,
AC² = BC² - AB²
$\Rightarrow$ AC² + AB² = BC²
$\therefore \triangle$ ABC is right angled at A (by using converse of Pythagoras theorem)
$\therefore$ $\angle$ A = 90°
Now, In $\triangle$ APC
$\angle$ PAC + $\angle$ C = 90° ........(i)
Similarly, $\angle$ B + $\angle$ C = 90° .........(ii)
from (i) and (ii), we get
$\angle$ B + $\angle$ C = $\angle$ PAC + $\angle$ C
$\Rightarrow$ $\angle$ B = $\angle$ PAC
In $\triangle$ BPA and $\triangle$ APC,
$\angle$ B = $\angle$ PAC (proved)
$\angle$ BPA = $\angle$ APC (each 90°)
$\therefore \quad \triangle B P A \sim \triangle A P C$
$\Rightarrow \quad \frac { \mathrm { PB } } { \mathrm { PA } } = \frac { \mathrm { PA } } { \mathrm { CP } }$
$\Rightarrow \quad \mathrm { PB } \times \mathrm { CP }= PA^2$
Hence proved.

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