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Standard IX

Mathematics

The Mid-Point Theorem

B 1 is a poin...

Question

$B_{1}$ is a point on the side AC of \Delta ABC and B_1 B is joined. A line is drawn through A parallel to B_1B, meeting CB produced in $A_{1}$ and another line is drawn through C parallel to $B_{1} B$ , meeting AB produced in $C_{1}$ prove that $\frac{1}{A A_{1}} + \frac{1}{C C_{1}} = \frac{1}{B B_{1}}$ [4 MARKS]

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Solution

Applying theorems: 2 Marks
Calculation: 2 Marks

Proof: $∠ A = ∠ A (c o m m o n) a n d ∠ A B B_{1} = ∠ A C_{1} C (C o r r e s p o n d i n g a n g l e s) ∴ Δ A B B_{1} \sim Δ A C_{1} C [A A S i m i l a r l y] ∴ \frac{B B_{1}}{C C_{1}} = \frac{A B_{1}}{A C} \dots (i) S i m i l a r l y Δ C B B_{1} \sim Δ C A_{1} A [A A S i m i l a r l y] ∴ \frac{B B_{1}}{C C_{1}} = \frac{A B_{1}}{A C} \dots (i i) A d d i n g (i) a n d (i i), w e g e t \frac{B B_{1}}{C C_{1}} + \frac{B B_{1}}{A A_{1}} = \frac{A B_{1}}{A C} + \frac{C B_{1}}{A C} B B_{1} [\frac{1}{C C_{1}} + \frac{1}{A A_{1}}] = \frac{A B_{1} + C B_{1}}{A C} B B_{1} [\frac{1}{C C_{1}} + \frac{1}{A A_{1}}] = \frac{A C}{A C} \Rightarrow \frac{1}{C C_{1}} + \frac{1}{A A_{1}} = \frac{1}{B B_{1}}$

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B1 is a point on the side AC of \Delta ABC and B_1 B is joined. A line is drawn through A parallel to B_1B, meeting CB produced in A1 and another line is drawn through C parallel to B1B, meeting AB produced in C1 prove that 1AA1+1CC1=1BB1 [4 MARKS]