# In figure, l || m and line segments AB, CD and EF are concurrent at point P. Prove that AEBF=ACBD=CEFD. - Mathematics

Sum

In figure, l || m and line segments AB, CD and EF are concurrent at point P. Prove that (AE)/(BF) = (AC)/(BD) = (CE)/(FD).

#### Solution

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

To prove: (AE)/(BF) = (AC)/(BD) = (CE)/(FD)

Proof: In ΔAPC and ΔBPD,

∠APC = ∠PBD   .....[Vertically opposite angles]

∠PAC = ∠PBD   .....[Alternative angles]

∴ ΔAPC ∼ ΔBPD   .....[By AA similarity criterion]

Then, (AP)/(PB) = (AC)/(BD) = (PC)/(PD)   ......(i)

In ΔAPE and ΔBPF,

∠APE = ∠BPF   ......[Vertically opposite angles]

∠PAE = ∠PBF   .....[Alternative angles]

∴ ΔAPE ∼ ΔBPF   .....[By AA similarity criterion]

Then, (AP)/(PB) = (AE)/(BF) = (PE)/(PF)   ......(ii)

In ΔPEC and ΔPFD,

∠EPC = ∠FPD   ......[Vertically opposite angles]

∠PCE = ∠PDF   .....[Alternative angles]

∴ ΔPEC ∼ ΔPFD  .....[By AA similarity criterion]

Then, (PE)/(PF) = (PC)/(PD) = (EC)/(FD)   ......(iii)

From equation (i), (ii) and (iii)

(AP)/(PB) = (AC)/(BD) = (AE)/(BF) = (PE)/(PF) = (EC)/(FD)

∴ (AE)/(BF) = (AC)/(BD) = (CE)/(FD)

Hence proved.

Concept: Similarity of Triangles
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#### APPEARS IN

NCERT Mathematics Exemplar Class 10
Chapter 6 Triangles
Exercise 6.4 | Q 13 | Page 75
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