Question
Answer:
:
Each point: 1 Mark
(a) In line segments AE and EC, point E is a common point
So, AE + EC = AC
(b) In part (a) we have proved that:
AE + EC = AC
⇒ AC – EC = AE
(c) For line segments BE and ED, point E is a common point.
So, BE + ED = BD
⇒ BE – BE = ED
(d) Also, BE = BE + ED
⇒ BD – DE = BE
(∵ line segment ED = line segment DE)
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:
Each point: 1 Mark
(a) In line segments AE and EC, point E is a common point
So, AE + EC = AC
(b) In part (a) we have proved that:
AE + EC = AC
⇒ AC – EC = AE
(c) For line segments BE and ED, point E is a common point.
So, BE + ED = BD
⇒ BE – BE = ED
(d) Also, BE = BE + ED
⇒ BD – DE = BE
(∵ line segment ED = line segment DE)
Was this answer helpful ?
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