Question

In the figure, PQRS is a trapezium in which PQ || RS. On PQ and RS, there are points E and F respectively such that EF intersects SQ at G. Prove that EQ x GS = GQ x FS.

Answer 1

PQRS is a trapezium

PQ || RS

E and F are points on PQ and RS which intersects diagonal SQ at G

Now, considering ΔGEQ and ΔGFS, we have

∠EGQ = ∠FGS     [vertically opposite angles]

∠EQG = ∠FSG     [alternate angles]  

∠GEQ = ∠GFS     [alternate angles]

∴ By AAA similarity,

ΔGEQ ~  ΔGFS

Since the corresponding sides of two similar triangles are proportional to each other.

∴ \(\frac{EQ}{FS} = \frac{GQ}{GS}\)

⇒ EQ x GS = GQ x FS  

Hence proved.

Answer 2

In ΔGEQ and ΔGFS

∠EGQ = ∠FGS   (vert. opp. angles)

∠EQG = ∠FSG  (alt. angles)

:.   ΔGES ~  ΔGFS   (AA similarity)

or,   EQ / FS = GQ GS

or,   EQ x GS = GQ x FS