Question

AB=AC,Show That AD>AB,where D is a mid point on CB​

Answer 1

Answer:

In ΔABC, ∠ABC=∠ACB=θ (let)  [as AB=AC ⇒ΔABC is isosceles]

Now, we know that , in an isosceles Δ, the base angles are acute as only one obtuse angle can exist in a Δ.

Now ∠ACD= 180°-∠ACB     [linear pair]

so, in ΔACD , ∠ACD is the largest angle.

so, ∠ACD > ∠ADC

so, from triangle inequality, side opposite to the larger angle is gt the side opposite to the smaller angle,

so, AD > AC

but, AC=AB   [given]

so, AD > AB

Quad Erat Demonstrandum.

Step-by-step explanation:

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