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In fig. , l || m and line segments AB, CD and EF are concurrent at point P. Prove tha

t $\frac{AE}{BF}=\frac{AC}{BD}=\frac{CE}{FD}$ .

Answers (1)

Given: l || m and line segments AB, CD and EF are concurrent at point P

To prove:-

$\frac{AE}{BF}=\frac{AC}{BD}=\frac{CE}{FD}$

Proof :In $\Delta APC \; and\; \Delta DPB$

$\angle APC=\angle DPB$ (vertically opposite angles)

$\angle PAC=\angle PBD$ (alternate angles)

and we know that if two angles of one triangle are equal to the two angles of another triangle then the two triangles are similar by AA similarity criterion

$\therefore \; \; \; \; \; \Delta APC\sim \Delta DPB$

Then, $\frac{AP}{BP}=\frac{AC}{BD}=\frac{PC}{PD}\; \; \; \; \; \; \; \; \; ....(1)$

In $\Delta APE\; and\; \Delta FPB$

$\angle APE=\angle BPF$ (vertically opposite angle)

$\angle PAE=\angle PBF$ (alternate angle)

$\therefore \; \; \; \; \Delta APE\sim \Delta FPB$ (by AA similarity criterion)

Then $\frac{AP}{PB}=\frac{AE}{BF}=\frac{PE}{PF}\; \; \; \; \; \; \; ...(2)$

In $\Delta PEC\; and\; \Delta PFD$

$\angle EPC=\angle FPD$ (vertically opposite angle)

$\angle PCE=\angle PDF$ (alternate angle)

$\therefore \; \; \; \; \Delta PEC\sim \Delta PFD$ (by AA similarity criterion)

Then $\frac{PE}{PF}=\frac{PC}{PD}=\frac{EC}{FD}\; \; \; \; \; \; \; ...(3)$

From equation (1), (2) and (3) we get

$\frac{AP}{BP}=\frac{AC}{BD}=\frac{AE}{BF}=\frac{PE}{PF}=\frac{EC}{FD}$

$\Rightarrow \frac{AE}{BF}=\frac{AC}{BD}=\frac{CE}{FD}$

Hence proved.

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