Vertices A and D are on the same side of BC. If AD is extended to intersect BC at E, show that
(i) ΔABD ≅ ΔACD
(ii) ΔABE ≅ ΔACE
(iii) AE bisects ∠A as well as ∠D
(iv) AE is the perpendicular bisector of BC.
Answers
Vertices A and D are on the same side of BC. If AD is extended to intersect BC at E, show that
(i) ΔABD ≅ ΔACD
(ii) ΔABE ≅ ΔACE
(iii) AE bisects ∠A as well as ∠D
(iv) AE is the perpendicular bisector of BC.
Step-by-step explanation:
The Congruence of triangles:
Two angles are congruent if the sides and the angles of the triangle are equal to the corresponding sides and angles of the other angle.
In Congruent Triangles corresponding parts are always equal and we denote it in short by CPCT : corresponding parts of Congruent Triangles
Criteria for congruence of triangles:
side angle side
Two Triangles are congruent if two sides and the included angle of a triangle are equal to the two sides and included angle of the the other triangle
side side side
If three sides of One triangle are equal to the three sides of another triangle then the two Triangles are said to be congruent.
Given,
ΔABC and ΔDBC are two isosceles triangles in which AB=AC & BD=DC.
To Prove:
(i) ΔABD ≅ ΔACD
(ii) ΔABE ≅ ΔACE
(iii) AE bisects ∠A as well as ∠D
(iv) AE is the perpendicular bisector of BC.
(i) In ΔABD and ΔACD,
AD = AD (Common)
AB = AC (given)
BD = CD (given)
Therefore, ΔABD ≅ ΔACD (by SSS congruence rule)
∠BAD = ∠CAD (CPCT)
∠BAE = ∠CAE
(ii) In ΔABE & ΔACE,
AE = AE (Common)
∠BAP = ∠CAP
(Proved above)
AB = AC (given)
Therefore,
ΔABP ≅ ΔACE
(by SAS congruence rule).
(iii)
∠BAD = ∠CAD (proved in part i)
∴, AE bisects ∠A.
also,
In ΔBPD and ΔCPD,
ED = PD (Common)
BD = CD (given)
BE = CE (ΔABE ≅ ΔACE so by CPCT.)
∴, ΔBED ≅ ΔCED (by SSS congruence rule.)
Thus,
∠BDE = ∠CDE( by CPCT.)
Hence, we can conclude that AE bisects ∠A as well as ∠D.
(iv)
∠BPD = ∠CED
(by CPCT as ΔBED ≅ ΔCED)
& BE = CE (CPCT)
also,
∠BED + ∠CED = 180° (BC is a straight line.)
2∠BED = 180°
∠BED = 90°
∴
AE is the perpendicular bisector of BC.
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