In triangle $A B C, P$ is a point in $A B$ such that $∠ A C P = ∠ A B C .$ If $A C = 9 c m, C P = 12 c m,$ and $B C = 15 c m;$ find $A P$ to closest integer.

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Solution

In $△ A C P$ and $△ A B C$ ,
$∠ A B C = ∠ A C P$ (Given)
$∠ B A C = ∠ P A C$ (Common)
$∠ A C B = ∠ A P C$ (Third angle)
Thus, $△ A C P \sim △ A B C$ (AAA rule)
Hence, $\frac{A P}{A C} = \frac{P C}{B C}$
$\frac{A P}{9} = \frac{12}{15}$
$A P = \frac{9 \times 12}{15}$
$A P = 7.2$ cm

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In triangle ABC,P is a point in AB such that ∠ACP=∠ABC. If AC=9cm,CP=12cm, and BC=15cm; find AP to closest integer.

In △ACP and △ABC,∠ABC=∠ACP (Given)∠BAC=∠PAC (Common)∠ACB=∠APC (Third angle)Thus, △ACP∼△ABC (AAA rule)Hence, APAC=PCBCAP9=1215AP=9×1215AP=7.2 cm

In triangle $A B C, P$ is a point in $A B$ such that $∠ A C P = ∠ A B C .$ If $A C = 9 c m, C P = 12 c m,$ and $B C = 15 c m;$ find $A P$ to closest integer.