Ramesh and Reena are playing with triangle ABC. Ramesh draws a line that bisects $$\angle BAC$$; this line cuts BC at D. Reena then extends AD to a point P. In response, Ramesh joins B and P. Reena then announces that BD bisects $$\angle PBA$$, hat a surprise! Together, Ramesh and Reena find that BD= 6 cm, AC= 9 cm, DC= 5 cm, BP=8 cm, and DP = 5 cm.
How long is AP?
Given:
BD= 6 cm, AC= 9 cm, DC= 5 cm, BP=8 cm, and DP = 5 cm.
Since AD is the angular bisector applying the angular bisector theorem we have :
$$\frac{AB}{BD}=\ \frac{AC}{CD}$$
Hence : Considering AB = x cm.
$$\frac{9}{5}=\ \frac{x}{6}$$
x = 10.8 cm.
Now since BD is the angular bisector for angle PBA we have :
Applyinh the internal angle bisector theorem :
$$\frac{PB}{PD}=\ \frac{BA}{AD}$$
Considering AD = y cm.
$$\frac{8}{5}\ =\ \frac{10.8}{y}$$
y = 6.75 cm.
AP = AD + DP.
= 6.75 + 5 = 11.75 cm
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