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prove isosceles triangle, given perpendicular line

\alpha \beta \gamma \theta \pi = \cdot \frac{\msquare}{\msquare} x^2 \sqrt{\square} \msquare^{\circ}
\:\:\:\bullet^{A} \angle \overline{AB} \sim \cong \angle \overline{AB} \overarc{AB}
\bigtriangleup \square \bigcirc \bigtriangleup S P \perpendicular \parallel
Given Prove Find

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