Question

∆ ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on same side of BC . If AD is extended to intersect BC at P show that
AP is the perpendicular bisector of BC

Answer 1

i)in trngle ABD and triangle ACD

AB=AC[GIVEN]

AD=AD[COMMON]

DB=CD[GIVEN]

=triangle ABD congruent to triangle ACD

= angle BAD=CAD [CPCT]-1

=angle BDA= CDA [CPCT] -2

(II) in triangle ABP and ACP

AB =AC

BAP=CAP-1

AP =AP

therefore traiangle ABP congruent ACP

=BP=CP[CPCT]-3

Now BP=CP

= angle BAD=CAD [CPCT]-1

=angle BDA= CDA [CPCT] -2

therefore ad bisects angle A as well as angle D