Truss bridges are formed with a structure of connected elements that form triangular structures to make up the bridge. Trusses are the triangles that connect to the top and bottom cord and two end posts. You can see that there are some triangular shapes as shown in the picture given below and these are represented as ΔABC, ΔACAD and ΔABE.
Based on the above situation, answer the following questions :
(i) If AB=CD and AD=BC, by which congruency rule ΔABC≅ΔCDA? (1)
(ii) If BE=AC and ∠EBA=∠BAC, by which congruency rule ΔEBA≅ΔCAB? (1)
(iii) (A) If AM⊥BC and AB=AC, prove that AM bisects BC. (2)
OR
(B) If A is the mid point of DE, BE||CA and AB||DC show that EB=AC. (2)
Doubt by Parth
Solution :
(i) AB=CD (Given)
BC=AD (Given)
AC=CA (Common)
Hence ΔABC≅ΔCDA (by SSS)
(ii) BE=AC (Given)
BC=AD (Given)
AC=CA (Common)
Hence ΔABC≅ΔCDA (by SSS)
(ii) BE=AC (Given)
∠EBA=∠BAC
AB=BA (common)
AB=BA (common)
Hence, ΔEBA≅ΔCAB (by SAS)
(iii)
Given : AM⊥BC
AB=AC
To prove : AM bisects BC
Proof :
In ΔAMB and ΔAMC
∠AMB=∠AMC (Each 90°)
AB=AC (Given)
AM=AM (Common)
AB=AC
To prove : AM bisects BC
Proof :
In ΔAMB and ΔAMC
∠AMB=∠AMC (Each 90°)
AB=AC (Given)
AM=AM (Common)
ΔAMB≅ΔAMC (By RHS)
BM=CM (By CPCT)
BM=CM (By CPCT)
Hence, AM bisects BC
OR
OR
Given : EA=DA
BE||CA
AB||DC
To Prove : EB=AC
BE||CA
AB||DC
To Prove : EB=AC
Proof :
In ΔEBA and ΔACD
∠AEB=∠DAC (Corresponding angles)
EA=AD (Given)
∠EAB=∠ADC (Corresponding angles)
ΔEBA≅ΔACD (By ASA)
EB=AC (By CPCT)
In ΔEBA and ΔACD
∠AEB=∠DAC (Corresponding angles)
EA=AD (Given)
∠EAB=∠ADC (Corresponding angles)
ΔEBA≅ΔACD (By ASA)
EB=AC (By CPCT)
Hence Proved ∎