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secondary 4 | E Maths
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Help asdfghjkl! Anyone can contribute an answer, even non-tutors.
pls help with 7a
Date Posted: 5 years ago
Views: 196
∠ TXA = ∠ XBA (angles in alternate segments)
∠ ZBA = 180° - ∠ XBA
(sum of angles on a straight line = 180°)
There is a cyclic quadrilateral ZBAY.
∠ ZBA = 180° - ∠ ZYA
( opposite angles of a cyclic quadrilateral add up to 180°)
From above , it can be deduced that
∠ ZYA = ∠ XBA = ∠ TXA
Since XAY is a straight line,
∠ ZYA = ∠ ZYX
∠ TXA = ∠ TXY
Thus, ∠ ZYX = ∠ TXY.
The two angles are alternate angles.
And hence TX // YZ
∠ ZBA = 180° - ∠ XBA
(sum of angles on a straight line = 180°)
There is a cyclic quadrilateral ZBAY.
∠ ZBA = 180° - ∠ ZYA
( opposite angles of a cyclic quadrilateral add up to 180°)
From above , it can be deduced that
∠ ZYA = ∠ XBA = ∠ TXA
Since XAY is a straight line,
∠ ZYA = ∠ ZYX
∠ TXA = ∠ TXY
Thus, ∠ ZYX = ∠ TXY.
The two angles are alternate angles.
And hence TX // YZ
thank you! :-)
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