Question

in triangle abc, f is a point bc. fp is drawn parallel to ca meeting ab in p and fe is drawn parallel to ba meeting ac at e. ep and cb when produced meet at d. prove that df square= db * dc

Answer 1

Step-by-step explanation:

Given in triangle abc, f is a point bc. fp is drawn parallel to ca meeting ab in p and fe is drawn parallel to ba meeting ac at e. ep and cb when produced meet at d. prove that df square= db * dc

• We need to prove df^2 = db x dc
• Given fp is parallel to ca and fe parallel to ba
• So fe is parallel to bp
• In triangle fde, we have bp is parallel to fe
• We have by basic proportionality theorem,
•  So db / df = dp / de ---------1
• In triangle edc we have fp is parallel to ce since fp is parallel to ca
• So by Basic proportionality theorem we have,
• So df / dc = dp / de -----------2
• So from equation 1 and 2 we get
•       So db / df = df / dc
• Or df^2 = db x dc

Reference link will be

https://brainly.in/question/2089382

Answer 2

Answer:

it will be the correct answer