Question

013. AB is a line segment & P is its mid-point. D &
E are points on the same side of AB such that
BAD ZABE& ZEPA - DPB. Show that
1 D) A DAP A A EBP
II) AD BE
C
OR
In an isosceles AABC with AB = AC, the bisector of
B & C intersects each other at O. Join A to O.
Show that
1) OBOC) AO bisects ZA​

Answer 1

Step-by-step explanation:

∠EPA=∠DPB

∠EPA=∠DPBSo, ∠APE+∠EPD=∠BPD+∠DPE

∠EPA=∠DPBSo, ∠APE+∠EPD=∠BPD+∠DPEHence,

∠EPA=∠DPBSo, ∠APE+∠EPD=∠BPD+∠DPEHence,∠APD=∠BPE ....(1)

∠EPA=∠DPBSo, ∠APE+∠EPD=∠BPD+∠DPEHence,∠APD=∠BPE ....(1)Opposite side of ∠APD is AD and opposite side of ∠DPE is BE,

∠EPA=∠DPBSo, ∠APE+∠EPD=∠BPD+∠DPEHence,∠APD=∠BPE ....(1)Opposite side of ∠APD is AD and opposite side of ∠DPE is BE,From equation (1),

∠EPA=∠DPBSo, ∠APE+∠EPD=∠BPD+∠DPEHence,∠APD=∠BPE ....(1)Opposite side of ∠APD is AD and opposite side of ∠DPE is BE,From equation (1),AD=BE ......(2)

∠EPA=∠DPBSo, ∠APE+∠EPD=∠BPD+∠DPEHence,∠APD=∠BPE ....(1)Opposite side of ∠APD is AD and opposite side of ∠DPE is BE,From equation (1),AD=BE ......(2)Hence proved.

∠EPA=∠DPBSo, ∠APE+∠EPD=∠BPD+∠DPEHence,∠APD=∠BPE ....(1)Opposite side of ∠APD is AD and opposite side of ∠DPE is BE,From equation (1),AD=BE ......(2)Hence proved.AP=PB ......(3)

∠EPA=∠DPBSo, ∠APE+∠EPD=∠BPD+∠DPEHence,∠APD=∠BPE ....(1)Opposite side of ∠APD is AD and opposite side of ∠DPE is BE,From equation (1),AD=BE ......(2)Hence proved.AP=PB ......(3)In △DAP and △EBP,

∠EPA=∠DPBSo, ∠APE+∠EPD=∠BPD+∠DPEHence,∠APD=∠BPE ....(1)Opposite side of ∠APD is AD and opposite side of ∠DPE is BE,From equation (1),AD=BE ......(2)Hence proved.AP=PB ......(3)In △DAP and △EBP,∠APD=∠BPE ....(1)

∠EPA=∠DPBSo, ∠APE+∠EPD=∠BPD+∠DPEHence,∠APD=∠BPE ....(1)Opposite side of ∠APD is AD and opposite side of ∠DPE is BE,From equation (1),AD=BE ......(2)Hence proved.AP=PB ......(3)In △DAP and △EBP,∠APD=∠BPE ....(1)AD=BE ......(2)

∠EPA=∠DPBSo, ∠APE+∠EPD=∠BPD+∠DPEHence,∠APD=∠BPE ....(1)Opposite side of ∠APD is AD and opposite side of ∠DPE is BE,From equation (1),AD=BE ......(2)Hence proved.AP=PB ......(3)In △DAP and △EBP,∠APD=∠BPE ....(1)AD=BE ......(2)AP=PB ......(3)

∠EPA=∠DPBSo, ∠APE+∠EPD=∠BPD+∠DPEHence,∠APD=∠BPE ....(1)Opposite side of ∠APD is AD and opposite side of ∠DPE is BE,From equation (1),AD=BE ......(2)Hence proved.AP=PB ......(3)In △DAP and △EBP,∠APD=∠BPE ....(1)AD=BE ......(2)AP=PB ......(3)Hence, △DAP≅△EBP by (S.A.S). congurency

∠EPA=∠DPBSo, ∠APE+∠EPD=∠BPD+∠DPEHence,∠APD=∠BPE ....(1)Opposite side of ∠APD is AD and opposite side of ∠DPE is BE,From equation (1),AD=BE ......(2)Hence proved.AP=PB ......(3)In △DAP and △EBP,∠APD=∠BPE ....(1)AD=BE ......(2)AP=PB ......(3)Hence, △DAP≅△EBP by (S.A.S). congurencyIt's proved.