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ICSE Class 10 Answered

In triangle ABC, angle ABC is equal to twice the angle ACB, and bisector of angle ABC meets the opposite side at point P. Show that: ¡) CB:BA=CP:PA ii) AB*BC=BP*CA
Asked by dibyamrao000 | 11 May, 2018, 20:18: PM
answered-by-expert Expert Answer
According to the question,
angle ABC = 2 angle ACB
Let angle ACB = y
angle ABC =  2 angle ACB = 2y
BP is bisector of angle ABC
angle ABP = angle PBC = y
Using angle bisector thereom
AB/BC = CP/PA
In ΔABC and ΔAPB,
angle ABC = angle APB          exterior angle property
angle BCP = angle ABP
ΔABC ≈ ΔAPB                       AA 
AB/BP = CA/CB
AB × BC = AC × BP
 
Answered by Sneha shidid | 15 May, 2018, 11:14: AM

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