Question

ABC is an isosceles triangle in which AB = AC and E and D are midpoints of sides AB and AC respectively. Prove CE = BD.

Answer 1

Answer:

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

Answer By

kirtii

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