∆ABC and ∆DBC are two isosceles triangle on the same base BC and vertical A and D are on same side of BC . if AD is extended to intersect BC at P show that----
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Answer:
Data : ∆ABC and ∆DBC are two isosceles triangles on the same base BC and vertices A and D are on the same side of BC. AD is extended to intersect BC at P.
To Prove:
(i) ∆ABD ≅ ∆ACD
(ii) ∆ABP ≅ ∆ACP
(iii) AP bisects ∠A as well as ∠D.
(iv) AP is the perpendicular bisector of BC
(v) AD is the angular bisector of ∠A.
Proof:
(i) In ∆ABD and ∆ACD,
AB = AC (data)
BD = DC (data)
AD is common.
S.S.S. Congruence rule.
∴ ∆ABD ≅ ∆ACD
(ii) In ∆ABP and ∆ACP,
AB = AC (data)
∠ABP = ∠ACP (Opposite angles)
∠BAP = ∠CAP (∵ ∆ABD ≅ ∆ACD proved) Now
ASA postulate. ∆ABP ≅ ∆ACP.
(iii) ∆BAD ≅ ∆CAD proved.
AP bisects ∠A.
In ∆BDP and ∆CDP,
BD = DC (data)
BP = PC (proved)
DP is common.
∴ ∆BDP ≅ ∆CDP (SSS postulate)
∴ ∠BDP = ∠CDP
∴ DP bisects ∠D.
∴ AP bisects ∠D.
(iv) Now, ∠APB + ∠APC = 180°
(Linear pair) ∠APB + ∠APB = 180°
2 ∠APB = 180
∴ ∠APB = 18021802
∴∠APB = 90°
∠APB = ∠APC = 90°
BP = PC (proved)
∴ AP is the perpendicular bisector BC.
(v) AP is the angular bisector of ∠A.
Angular bisector of ∠A is aD, because AD, AP are in one line.
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To Show :
∆ABD ≅ ∆ACD.
∆ABP ≅ △ACP
AP bisects ∠A as well as ∠D
AP is perpendicular bisector of BC .
Proof :
In △ABD AND △ACD
AB = AC [ GIVEN]
AD = AD [ COMMON ]
BD = CD [ GIVEN ]
Therefore, ∆ABD ≅ ∆ACD.[SSS]
2. In △ABP and △ ACP
AB = AC [ GIVEN ]
∠BAP = ∠CAP [ C.P.C.T ]
AP = AP [ COMMON ]
Therefore , △ABP ≅ △ACP [ SAS ]
Other parts referred as attachment..
