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2014-09-02T17:23:52+05:30
AB = a metres, CD = b metres
Distance between them is p metres
O is the point of intersection of BC and AD.
OL is perpendicular to AC
Let OL = h metres
Triangle ABC ~ triangle LOC as angles CAB and CLO are right angles and angle C is common.

 \frac{CA}{CL}= \frac{AB}{LO}
\Rightarrow  \frac{p}{x} =  \frac{a}{h}  \Rightarrow x =  \frac{ph}{a}         ...(1)

Triangle ALO ~ triangle ACD as angles ALO and ACD are right angles and angle A is common.

 \frac{AL}{AC}= \frac{OL}{CD}
\Rightarrow \frac{y}{p} = \frac{h}{b} \Rightarrow y = \frac{ph}{b} ...(2)

From equations (1) and (2)

x+y =  \frac{ph}{a} +  \frac{ph}{b}

p = ph(\frac{1}{a}+ \frac{1}{b})

h(\frac{a+b}{ab})  = 1

h =  \frac{ab}{a+b}

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2014-09-02T20:31:26+05:30

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Draw a line horizontally parallel to BC, meeting DC at F and AB at E.

Triangles DFO and  OLB are similar, as angle DFO = angle OLB = 90deg. and angle DOF = angle OBL as OF || BL.
So,  OL/BL = DF/OF   and  OL = FC
also, OF = LC.

DC = DF+FC = OL*OF/BL + OL = OL [OF+BL] / BL
     = OL * [CL + BL]/BL = OL * BC / BL

a = OL * BC/BL
BL = OL*BC/a    -- eq 1

Triangles AEO and OLC are similar, as angle AEO = OLC = 90 deg and angle AOE = angle OCL, as  OE || CL.
So, AE/OE = OL/CL     and    OE = BL and BE = OL

so  AB = AE + BE = OL*OE/CL + OL = OL [ OE+CL]/ CL
     b = OL * [BL+CL]/CL = OL * BC / CL

    CL = OL*BC/b    -- eq 2

add BL and CL from equation 1 and 2

BC= BL+CL = OL*BC [ 1/a + 1/b ] = OL*BC [ (a+b)/ab ]
So, canceling BC on both sides we get,

OL = ab / (a+b)






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