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the area of an isosceles triangle also, prove it.

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Please diagram. Base is BC = a,  Two Sides  b = c ie., AB =  AC.  and  Angle B = angle C  : Isosceles triangle.

The perpendicular bisector AD bisects BC, because AB=AC. AD is perpendicular.

AD^2 + BD^2 = AB^2, \ \ \ Pythagoras\ law \\ AD^2 = b^2 - (\frac{BC}{2})^2 = \frac{1}{4} (4b^2 - a^2) \\ AD = \frac{1}{2} \sqrt{4b^2-a^2} \\ \\ Area\ of\ ISOSCELES\ Triangle\ =\ \frac{1}{2} base\ *\ altitude\ \\ = \frac{1}{2} * a * \frac{1}{2} \sqrt{4b^2 - a^2} \\ \\ = \frac{1}{4}  a\ \sqrt{4b^2 - a^2} \\

Another formula:

Sin\ B = \frac{Height}{b},\ \ \ Height = b\ Sin\ B \\ \\ Area\ =\ \frac{1}{2}\ a\ b\ Sin\ B \\ \\
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reacd the first line of answer.
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