A ABC and A DBC 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). IfAD is extended
to intersect BC at P. show that
(1) A ABDA ACD
(i) A ABP=AACP
B
(iii) AP bisects Z A as well as Z D.
actor of BC
Fig. 7.39
Answers
(i) In △ABD and △ACD,
AB = AC (since △ABC is isosceles)
AD = AD (common side)
BD = DC (since △BDC is isosceles)
Δ ABD ≅ Δ ACD (SSS test of congruence )
∴∠BAD = ∠CAD i.e. ∠BAP = ∠PAC -----------(1) [c.a.c.t]
(ii) In △ABP and △ACP,
AB = AC (since △ABC is isosceles)
AP = AP (common side)
∠BAP = ∠PAC from (1)
△ABP ≅△ACP (SAS test of congruence)
∴ BP = PC --------–--------(2). [c.s.c.t]
∠APB = ∠APC
(iii) Since △ABD ≅ △ACD
∠BAD = ∠CAD from (1)
So, AD bisects ∠A
i.e. AP bisects ∠A. ---------------(3)
In △BDP and △CDP,
DP = DP (common side)
BP = PC from (2)
BD = CD (since △BDC is isosceles)
△BDP ≅△CDP (SSS test of congruence)
∴∠ BDP = ∠CDP ....c.a.c.t.
∴ DP bisects ∠D
So, AP bisects ∠D ---------------(4)
From (3) and (4),
AP bisects ∠A as well as ∠D.
(iv) We know that
∠APB + ∠APC = 180∘ (angles in linear pair)
Also, ∠APB = ∠APC from (2)
∴ ∠APB = ∠APC = 180°/2 = 90∘
BP = PC and ∠APB = ∠APC = 90∘
Hence, AP is perpendicular bisector of BC.
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