Question

In isosceles triangle $ABC$, sides $AB$ and $AC$ are equal. If midpoint $D$ lies in base $BC$ and point $E$ lies on $BC$ produced (BC being produced through vertex C), then :

• A. $AD>AE$
• B. $AE>AD$
  
• C. $AC>AE$
• D. $AD>AC$

Answer 1

Answer: (B)

$AE>AD$

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