1. Using basic proportionality theorem, prove that a line drawn through the mid point of one side of a triangle parallel to another side bisects the third side.

Proof: Let D is the mid point of AB and DE is parallel to BC.

To prove: E is the mid point of AC.

Proof: Since DE is parallel to BC, therefore by basic proportionality theorem \frac{AB}{AD}=\frac{AC}{AE} \frac{AB}{AD}-1=\frac{AC}{AE}-1 \RightarrowCA x CA = CB x CD

\RightarrowCA2 = CB x CD proved


4. ABC is a right angled triangle, right angle at B and BD is perpendicular to AC. If AB = 6, BC = 8 and AC = 10, then BD = ?


\angleC = \angleC

\angleB = \angleD

Hence \triangle ABC is similar to BDC

Therefore, \frac{AB}{BD}=\frac{AC}{CD} \frac{6}{BD}=\frac{10}{6}

So BD = 3.6


5. ABC is a triangle  in which DE is parallel to BC. If AD = 4, BD = 3, AE = 12, find the value of AC.

Solution: Since DE is parallel to BC, therefore  \frac{AD}{BD}=\frac{AE}{CE} \frac{4}{3}=\frac{12}{x}

x = 9

so AC = 12 + 9 = 21


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