# a squar ABCD in which AC =BE when BC produced .A is joined to E prove that FG=GE when AE intersect BD at F and CD at G

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+1 vote
by Mature (229 points)
by Basic (25 points)
Ok, to prove FG=GE, produce a line l || BC, such that it meets AB in T and DG in K,
Now side of the square = a
Clearly, ∆FKG ~ ∆ECG

(note : "~" is stands for similar triangles in which two triangles have all are corresponding angles equal and the very common properties of similar triangles is that the corresponding sides are all have same ratio)

now to prove the proposition given, we need to prove
∆FKG ≅ ∆ECG, So for that it is enough to show any one corresponding side is equal to that of another.
Let's try with FK and CE.
These things are easily seen :
1.  AB=BC=CD=DA= a,
2.  BD= BE = √2a
3.  AE = √3a
4. CE = BE - BC = (√2-1)a

Now see, ∆ABE ~ ∆ GCE
so, AB/BE = GC/CE
CG = (1-1/√2)a
So DG = a - CG = a/√2 ........ (5)

Now look at ∆AFB ~ ∆GFD
So, AF/FG = AB/GD
put the value of GD from (5) in above to get
AF/FG = √2
Now, ∆ATF ~ ∆GKF
So, AF/TF = GF/KF
=> AF/FG = TF/KF
=> TF/KF = √2
So, (TF + KF)/KF = (√2 + 1)....... (6)
TF + KF = a
So (6) => FK = a/(√2+1) = (√2-1)a
which is equal to CE ... (Look at the 4th point above)
So, ∆FKG ≅ ∆ECG
=> FG = GE                     (proved)

+1 vote
+1 vote