Question

In the given figure,AB ll CD and a transversal t cut them at E and F respectively . If EG and FG are the bisectors of BEF and EFD respectively , prove that EGF = 90°​

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Answer 1

Step-by-step explanation:

We know that AB || CD and t is a transversal cutting at points E and F From the figure we know that ∠BEF and ∠DFE are interior angles So we get ∠BEF + ∠DFE = 180°

Dividing the entire equation by 2 we get (1/2)∠BEF + (1/2) ∠DFE = 90°

According to the figure the above equation can further be written as ∠GEF + ∠GFE = 90° ……. (1) According to the △ GEF We can write ∠GEF + ∠GFE + ∠EGF = 180°

Based on equation (1) we get 90° + ∠EGF = 180°

On further calculation ∠EGF = 180° – 90° By subtraction ∠EGF = 90°

Therefore, it is proved that ∠EGF = 90°

Answer 2

$\huge\mathfrak\pink{Solution:-}$

$\sf\red{We \ know \ that}$

• AB || CD and t is a transversal cutting at points E and F.

From the figure we know that ∠BEF and ∠DFE are interior angles.

$\sf\green{So \ we \ get}$

• ∠BEF + ∠DFE = 180°

Dividing the entire equation by 2 we get

• (1/2)∠BEF + (1/2) ∠DFE = 90°

According to the figure the above equation can further be written as

• ∠GEF + ∠GFE = 90° ……. (1)

According to the △ GEF

$\sf\purple{We \ can \ write}$

• ∠GEF + ∠GFE + ∠EGF = 180°

Based on equation (1) we get

• 90°+ ∠EGF = 180°

$\sf\pink{On \ further \ calculation}$

• ∠EGF = 180° – 90°

$\sf\red{By \ subtraction}$

• ∠EGF = 90°

Therefore, it is proved that ∠EGF = 90°