Question

The diagonals AC and BD of a cyclic quadrilateral ABCD intersect at E. From the point E, the perpendicular GE is drawn on AB. If GE is producebd to meet DC at F , then prove that DF = FC .

Answer 1

DF = FC , if The diagonals AC and BD of a cyclic quadrilateral ABCD intersect at E at right angles  

Step-by-step explanation:

Correct Question is : AC and BD intersect in E at right angles

.

We have ∠AEB=90∘ (∵the diagonals are at right angles)

⇒∠1  +  ∠2 = 90∘

∠1 = 90∘−∠2   Eq (1)

∠EGB=90∘

In ΔEGB, we have

∠2 + ∠4 = 90∘

⇒∠4=90∘−∠2   Eq.(2)

From Eq(1) and Eq (2) , we get

∠1=∠4

∠1=∠3 (vertically opposite angles)

Hence ∠3=∠4

Also, ∠4=∠5 (anglesby same chord AD in same arc segment)

=> ∠3=∠5

=> FC = EF  (sides opposite to equal angles of a triangle are equal)

Similarly we can show that  DF = EF

Hence from both

FC=DF

=> DF = FC

QED

Proved

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Answer 2

Step-by-step explanation:

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