Page 11 - Ans Sub

13. In the given figure, AB || CD and a transversal t cuts them at E and F respectively.
If EG and FG are the bisectors of ∠BEF and ∠EFD respectively, prove that ∠EGF
= 90 0

Solution:
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 co-interior angles.
So, we get
0
∠BEF + ∠DFE = 180
1
1 ∠BEF + ∠DFE = 90
0
2 2
0
∠GEF + ∠GFE = 90 ……. (1)
Now, in △ GEF,
0
∠GEF + ∠GFE + ∠EGF = 180
0
0
90 + ∠EGF = 180
0
0
∠EGF = 180 - 90
0
∠EGF = 90
0
Therefore, it is proved that ∠EGF = 90 .
0
14. In the given figure, AB || CD. Prove that p + q – r = 180 .

Solution:
Draw a line KH passing through the point F which is parallel to both AB and CD
We know that KF || CD and FG is a transversal
From the figure we know that ∠KFG and ∠FGD are alternate angles
So, we get
∠KFG = ∠FGD = r ……. (1)
We also know that AE || KF and EF is a transversal
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