My sum: Based on angles, angle bisector, cyclic quadrilateral, circle:
In triangle ABC /_ angle ABC (i.e. /_ ABC) equal to 90°. Ray BD is bisector of angle ABC meeting AC at D.
If ray DP is angle bisector of angle ADB and ray DQ is angle bisector of angle BDC.
Prove that:
(1) DA/DC = AB/BC
(2) (AP/PB) x (BQ/QC) = DA/DC = AB/BC
(3) Find measure of angle PDQ.
(4) Prove that quadrilateral BPDQ is cyclic.
(5) Prove that if a circle is drawn through these points i.e. B,P,D and Q, PQ will be diameter of the circle.
(6) If PQ and DB intersect at X,
prove that QX/XP = QB/BP.
(7) Hence prove that, (AP/QC) x (QX/XP) = DA/DC.
In triangle ABC /_ angle ABC (i.e. /_ ABC) equal to 90°. Ray BD is bisector of angle ABC meeting AC at D.
If ray DP is angle bisector of angle ADB and ray DQ is angle bisector of angle BDC.
Prove that:
(1) DA/DC = AB/BC
(2) (AP/PB) x (BQ/QC) = DA/DC = AB/BC
(3) Find measure of angle PDQ.
(4) Prove that quadrilateral BPDQ is cyclic.
(5) Prove that if a circle is drawn through these points i.e. B,P,D and Q, PQ will be diameter of the circle.
(6) If PQ and DB intersect at X,
prove that QX/XP = QB/BP.
(7) Hence prove that, (AP/QC) x (QX/XP) = DA/DC.
