Question

4. Q is a point on the line BD dividing the segment internally. AB, PO and CD are drawn
perpendicular to BD. If AB = a, PQ = b and CD = c, then
1)1/a+1/b=1/c
2)1/a+1/c=1/b
3)1/a-1/b=1/c
4)1/b+1/c=1/a
answer me fast..............
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Answer 1

Given:

Q is a point on the line BD diving it internally

AB ⊥ BD, PQ ⊥ BD & CD ⊥ BD

AB = a, PQ = b and CD = c

To find:

The relation between a, b & c.

Solution:

In ΔABD & ΔPQD,

∠ADB = ∠PDQ ..... [common angle]

∠ABD = ∠PQD = 90° ...... [∵ AB⊥BD & PQ⊥BD]

∴ ΔABD ~ ΔPQD ..... [By AA similarity]

Since the corresponding sides of two similar triangles are proportional to each other.

∴ $\frac{AB}{PQ} = \frac{BD}{QD}$

substituting AB = a & PQ = b

$\implies \frac{a}{b} = \frac{BD}{QD}$

$\implies \frac{b}{a} = \frac{QD}{BD}$ ..... (i)

In ΔCDB & ΔPQB,

∠CBD = ∠PBQ ..... [common angle]

∠CDB = ∠PQB = 90° ...... [∵ CD⊥BD & PQ⊥BD]

∴ ΔCDB  ~ ΔPQB ..... [By AA similarity]

Similarly, we get

∴ $\frac{CD}{PQ} = \frac{BD}{QB}$

substituting CD = c & PQ = b

$\implies \frac{c}{b} = \frac{BD}{QB}$

$\implies \frac{b}{c} = \frac{QB}{BD}$ ..... (ii)

Now, adding (i) & (ii), we get

$\frac{b}{a} + \frac{b}{c} = \frac{QD}{BD} + \frac{QB}{BD}$

$\implies b [\frac{1}{a} + \frac{1}{c}] = \frac{QD + QB}{BD}$

since BD = QD + QB

$\implies b [\frac{1}{a} + \frac{1}{c}] = \frac{BD}{BD}$

$\implies b [\frac{1}{a} + \frac{1}{c}] =1$

$\implies\bold{ \frac{1}{a} + \frac{1}{c} = \frac{1}{b}}$ ← option (2)

Thus, the final answer is $\boxed{\underline{\underline{ \frac{1}{a} + \frac{1}{c} = \frac{1}{b}}}}}.$

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Answer 2

Step-by-step explanation:

1/a+1/c=1/b............