In triangle ABC,point D lies on BC
Such that AD=AC. prove that AB>AD
Answers
In triangle ABC, point D lies on BC such that AD=AC then it is proved that AB>AD.
Step-by-step explanation:
It is given that,
ABC is a triangle where D is a point on BC such that
AD = AC ....... (i)
From (i), we get
∠ACD = ∠ADC ...... [∵ angles opposite to equal sides are also equal in length] ...... (ii)
Now, from the figure attached below, we can see that ∠ADC is an exterior angle to ΔABD.
We know that exterior angle to a triangle is the sum of interior opposite angles of that triangle.
∴ ∠ADC = ∠ABD + ∠BAD ...... (iii)
So, we can also write from (iii),
∠ADC > ∠ABD
⇒ ∠ACD > ∠ABD ...... [substituting from (ii)]
∵ if in a triangle two angles are not equal, then the side opposite to the greater angle is longer than the side opposite to the smaller angle.
∴ AB > AC
from (i) AD = AC
∴ AB > AD
Hence proved
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