Mathematics
The given figure shows two isosceles triangles ABC and DBC with common base BC. AD is extended to intersect BC at point P. Show that :
(i) △ABD ≡ △ACD.
(ii) △ABP ≡ △ACP.
(iii) AP bisects ∠BDC.
(iv) AP is perpendicular bisector of BC.
Triangles
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Answer
Given: ABC and DBC with common base BC. AD is extended to intersect BC at point P.
(i) To prove: △ABD ≡ △ACD.
Proof: In △ABD and △ACD,
AB = AC (△ABC is an isosceles)
AD = AD (Common side)
BD = DC (△BDC is an isosceles)
By SSS congruency criterion,
Hence, △ABD ≅ △ACD.
(ii) To prove: △ABP ≡ △ACP.
Proof: From (i), △ABD ≅ △ACD
By corresponding parts of congruent triangles,
∠BAD = ∠CAD
i.e., ∠BAP = ∠PAC
In △ABP and △ACP,
AB = AC (△ABC is an isosceles)
AP = AP (Common side)
∠BAP = ∠PAC (proved above)
By SAS congruency criterion,
Hence, △ABP ≅ △ACP.
(iii) To prove: AP bisects ∠BDC.
Proof: From (ii), △ABP ≅ △ACP
By corresponding parts of congruent triangles,
⇒ BP = PC
In △BDP and △CDP,
DP = DP (common side)
BP = PC (proved above)
BD = DC (△BDC is an isosceles)
By SSS congruency criterion,
Hence, △BDP ≅ △CDP.
By corresponding parts of congruent triangles,
∠BDP = ∠CDP
Hence, AP bisects ∠BDC.
(iv) To prove: AP is perpendicular bisector of BC.
Proof: We know that ∠APB and ∠APC form linear pair.
⇒ ∠APB + ∠APC = 180°
Also, ∠APB = ∠APC
∵ ∠APB = ∠APC = = 90°
BP = PC and ∠APB = ∠APC = 90°
Hence, AP is perpendicular bisector of BC.
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