Question

In the above picture ∆ ABC is an isosceles right angled triangle whose <C = 90°. [<ADC = <CDB = 90°]

Prove that : CD = AD in ∆ ACD​

Answer 1

It is so easy but important question.

Here ,

In ∆ABC angle C = 90 and angle A and angle B are equal because AC = BC .

{ sides opposite to equal angles are equal. }

AD = BD is given angle ACD = angle BCD.

{ angles opposite to equal sides are equal.}

angle ACD = angle BCD { proved above }

CD = BD { sides opposite to equal angles are equal. }

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