Question

A
M
the arms of the angle.
Hint: Take any point P on bisector BD of ZABC. Draw PM 1 on AB and
PN 1 BC. Then prove APBM =APBN (by ASA criterion of congruence).
At last, prove PM = PN (as corresponding parts of congruent triangles
2
с
are equal)]
Prove that any point on the bisector of an angle is equidistant from​

Answer 1

Given:ABD=CBD

TO PROVE: AD=CD

CONSTRUCTION: DRAW AD PERPENDICULAR

AB AND CD PERPENDICULAR

BC

PROOF: In triangle ABD and triangle CBD

ABD=angle CBD.......(given)

seg BD=seg BD ......... (common side)

DAB=DCB..........(each 90°)

thereforeADB=CBD........(S-A-S test)

therefore seg AD=seg CD.......(c.s.c.t)

______(hence proved)