in isosceles triangle ABC sides AB and AC are equal. If point D lies in base BC and point E lies on BC produced (BC being produced through vertex C), prove that: (1) AC>AD
(2)AE>AC
(3)AE>AD
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In ∆ABC,
We have:
AB = AC⇒∠ACB = ∠ABC
[Angles opposite to equal sides are equal] ....(1)
We know that exterior angle of a ∆ is always greater than each of the interior opposite angle.
In ∆ADC,
∠ADB > ∠ACD
⇒∠ADB > ∠ABD [Using (1)]
⇒AB > AD
[Sides opposite to greater angle is longer]
⇒AC > AD [As, AB = AC] .....(2)
We know that exterior angle of a ∆ is always greater than each of the interior opposite angle.
In ∆ACE,
∠ACD > ∠AEC
⇒∠ABE > ∠AEB [Using (1)]
⇒AE > AB
[Sides opposite to greater angle is longer]
=> AE > AC [As, AB=AC].......... (3)
From (2) & (3) we get,
AE > AD
We have:
AB = AC⇒∠ACB = ∠ABC
[Angles opposite to equal sides are equal] ....(1)
We know that exterior angle of a ∆ is always greater than each of the interior opposite angle.
In ∆ADC,
∠ADB > ∠ACD
⇒∠ADB > ∠ABD [Using (1)]
⇒AB > AD
[Sides opposite to greater angle is longer]
⇒AC > AD [As, AB = AC] .....(2)
We know that exterior angle of a ∆ is always greater than each of the interior opposite angle.
In ∆ACE,
∠ACD > ∠AEC
⇒∠ABE > ∠AEB [Using (1)]
⇒AE > AB
[Sides opposite to greater angle is longer]
=> AE > AC [As, AB=AC].......... (3)
From (2) & (3) we get,
AE > AD
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