Question

In a triangle Abc , a line is drawn parallel to bc to meet ab at d and ac at e. DC and BE meet at X. Prove AD/AB = EX/XB

Answer 1

Answer:

hey here is your proof

pls mark it as brainliest

refer the reference diagram uploaded above

$now \: considering \: △DXE \: \: and \: \: △BXC \\ ∠DXE = ∠BXC \: \: \: \: \: \: \: (vertically \: opposite \: angles) \\ since \: DE \: || \: BC \\ ∠DEX=∠XBC \: \: \: \: \: \: \: (alternate \: angles \: theorem) \\ \\ △DXE \: \: is \: similar \: to \: △BXC \: \: \: \: \: \: \: (AA \: test \: of \: similarity) \\ \\ hence \: then \\ \frac{EX}{XB} = \frac{DE}{BC} \: \: \: \: \: \: \: (c.s.s.t) \: \: \: \: \: (2)$

$so \: from \: (1) \: and \: (2) \\ we \: get \\ \\ \frac{AD}{AB} = \frac{EX}{XB} \\ thus \: proved$

$To \: prove = \\ \frac{AD}{AB} = \frac{EX}{XB} \\ \\ so \: here \: BC \: || \: DE \\ so \: considering \: △ADE \: \: and \: \: △ABC \\ ∠DAE= ∠BAC \: \: \: \: \: \: \: \: (common \: angle) \\ since \: BC \: || \: DE \\ ∠ ADE=∠ABC \: \: \: \: \: (corresponding \: angles) \\ \\ △ADE \: \: is \: similar \: to \: △ABC \: \: \: \: \: \: \: (AA \: test \: of \: similarity) \\ hence \: then \\ \frac{AD}{AB} = \frac{DE}{BC} \: \: \: \: \: \: \: (c.s.s.t) \: \: \: \: \: \: \: (1)$