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`DeltaA B C`and `DeltaD B C`are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC (see Fig. 7.39). If AD is extended to intersect BC at P, show that (i) `\ DeltaA B D~=DeltaA C D`(ii) `DeltaA B P~=Delta

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Given : `DeltaABC` and `DeltaDBC` are two isosceles triangles having common base BC, such that AB=AC and DB = DC.
To prove : (i) `DeltaABD cong DeltaACD`
(ii) `DeltaABP cong DeltaACP`
(iii) AP bisects `angleA` as well as `angleD`.
(iv) AP is the perpendicular bisector of BC.
Proof : (i) In `DeltaABD` and `DeltaACD`, we have
AB=AC (given)
BD=CD (given)
and AD=AD (common)
`:. DeltaABD cong DeltaACD` (by SSS congruence axiom)
(ii) In `DeltaABP` and `DeltaACP`, we have
`AB=AC` (given)
`angle1=angle2" "( :' DeltaABD cong ACD)`
and `AP=AP` (common)
`:. DeltaABP cong DeltaACP` (by SAS congruence axiom)
(iii) `angle1=angle2 ( :' DeltaABD cong DeltaACD)`
`implies" "AP` bisects `angleA`
`angleADB=angleADC ( :' DeltaABD cong DeltaACD)`
`implies" "180^(@)- angleADB=180^(@)-angleADC`
`implies" "angleBDP=angleCDP` (by linear pair axiom)
`implies" "AP` bisects `angleD`.
(iv) `BP=CP" "( :' DeltaABP cong DeltaACP)`
and `angle3=angle4` ...(1)
But `angle3+angle4=180^(@)` (linear pair)
`:. angle3+angle3=180^(@)" "implies" "angle3=90^(@)` [from (1)]
`implies angle3=angle4=90^(@)`
Hence, AP is the perpendicular bisector of BC.
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