# ABC and DBC are two triangles on the same base BC.and on the same side of BC with angle A=angleD=90 degrees. if CA and BD meet each other at E. Show that AE*EC=BE*ED

SHAIK AASIF AHAMED
9 years ago
Hello student,
Given : In triangles ABC and DBC
angle A=angleD=90 degrees
AC and BD intersect each other at E
To Prove : AE x EC = ED x BE
Proof : In triangles AEB and EDC,
and AEB = DEC..............(Vertically opposite angles)
triangle ABE ~ EDC
(EB/EC) = (AE/DE)
EB x DE = EC x AE
Hence, AE x EC = ED x BE
S. Haris Ahmed Irfan
29 Points
9 years ago
Please can you show the figure
SHAIK AASIF AHAMED
9 years ago
Hello student,
Given : In triangles ABC and DBC
angle A=angleD=90 degrees
AC and BD intersect each other at E
To Prove : AE x EC = ED x BE
Proof : In triangles AEB and EDC,
and AEB = DEC..............(Vertically opposite angles)
triangle ABE ~ EDC
(EB/EC) = (AE/DE)
EB x DE = EC x AE
Hence, AE x EC = ED x BE
gauri
23 Points
9 years ago
in triangle AEB n DEC
BY AA Triangles r similar
THEN AE\ED=BE\CE
Then AE*CE=BE*ED
4 years ago
Dear student,
Given : In triangles ABC and DBC
angle A=angleD=90 degrees
AC and BD intersect each other at E
To Prove : AE x EC = ED x BE
Proof : In triangles AEB and EDC, and AEB = DEC..............(Vertically opposite angles)
triangle ABE ~ EDC
(EB/EC) = (AE/DE)
EB x DE = EC x AE
Hence, AE x EC = ED x BE

Thanks and regards,
Kushagra