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In Figure, line l is the bisector of ...

In Figure, line `l` is the bisector of angle `A\ a n d\ B` is any point on `ldotB P\ a n d\ B Q` are perpendiculars from `B` to the arms of `Adot` Show that: ` A P B\ ~= A Q B` BP=BQ or B is equidistant from the arms of `/_A`

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`(##IX_07_E01_05_s01##)`
Given: l is the bisector of an angle `∠A` and `BP ⊥ AP` and `BQ ⊥ AQ`
To Prove: `ΔAPB ≅ ΔAQB `and `BP = BQ`
i) We can show two triangles APB and AQB are congruent by using AAS congruency rule and then show that the corresponding parts of congruent triangles will be equal. ...
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