Question

in a figure ab ,dc and ef perpendicular to bc such that ab=p , dc=q ,ef= r, bf=a and cf=b. prove that 1/p+1/q=1/r.

Answer 1

Answer:

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Answer 2

$\frac{1}{p} + \frac{1}{q} = \frac{1}{r}$

GIVEN

AB ⏊ BC

DC ⏊ BC

EF ⏊ BC

AB = p

DC = q

EF = r

BF = a

CF = b

TO FIND

To prove

$\frac{1}{p} + \frac{1}{q} = \frac{1}{r}$

SOLUTION

We can simply solve the above problem as follows,

In ΔABC and Δ EFC

Since AB and EF is perpendicular to a common line BC

Therefore

∠CFE = ABC = 90°

And,

∠FCE = BCA (Common angle)

Therefore,

By Angle-Angle similarity criterion

ΔEFC ≈ ΔABC

Therefore,

$\frac{EF}{AB \:} = \frac{FC}{BC}$

Putting the values in above,

r/p = b/a+b (equation 1 )

Now,

In Δ BDC and Δ BFE

∠BCD = ∠BFE

And,

∠CBD = ∠FBE (Common angle)

By Angle-Angle similarity criterion,

Δ BDC ≈ Δ BFE

Therefore,

EF/DC = BF/BC

Putting the values of EF, DC,BF and BC

r/q = a/a+b (equation 2 )

Adding equation 1 and 2

$\frac{r}{p} + \frac{r}{q} = \frac{a}{a + b} + \frac{b}{a + b \: }$

$\frac{r}{p} + \frac{r}{q} = 1$

taking 'r' as common in LHS

$r( \frac{1}{p} + \frac{1}{q} ) = 1$

we can also write it as -

$\frac{1}{p} + \frac{1}{q} = \frac{1}{r}$

Hence, proved.

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