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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.

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

From equations (1) and (2)

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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)*

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

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

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,