In the given figure, AD is the bisector of the exterior A of ABC . Seg AD intersect the side BC produced in D. Prove that : BD AB = CD AC (Hint : Draw seg CE seg AD )
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Apply Basic Proportionality Theorem
Math
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Step-by-step explanation:
ABC is a triangle; AD is the exterior bisector of
Given:
\angle A
\angle A
and meets BC
produced at D; BA is produced to F.
\frac {BD}{CD} = \frac {AB}{AC}
\frac {BD}{CD} = \frac {AB}{AC}
To prove:
Construction: Draw CE||DA to meet AB at E.
Proof: In ABC. CE||AD cut by AC.
\angle CAD = \angle ACE
\angle CAD = \angle ACE
(Alternate angles)
Similarly CE || AD cut by AB
\angle FAD = \angle AEC
\angle FAD = \angle AEC
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