Question

in figure ad perpendicular to bc and be=ec. if DC =3/4 bc and de =1/4 bc to prove bc square= 2ac square -2ae square​

Answer 1

Answer:

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Answer 2

Step-by-step explanation:

We are given

here

AD

ˉ

⊥

BC

ˉ

& DB=3CD−−−(1)

so, BC=DB+CD

BC=3CD+CD

so, BC=4CD−−−(2)

Hence it given that,

AB

ˉ

is perpendicular to

BC

ˉ

by the pythagoras theorem

In both triangle △ABD & △ACD

AB

2

=AD

2

+BD

2

& AC

2

=AD

2

+CD

2

∴ AD

2

=AB

2

−DB

2

& AD

2

=AC

2

−CD

2

Compare both AD

2

with each other we get,,

AB

2

−DB

2

=AC

2

−CD

2

AB

2

=AC

2

+DB

2

−CD

2

AB

2

=AC

2

+(3 CD)

2

−CD

2

(from (1))

AB

2

=AC

2

+9 CD

2

−CD

2

AB

2

=AC

2

+8 CD

2

AB

2

=AC

2

+8(

4

BC

)

2

(from (2))

2AB

2

=2 AC

2

+BC

2

(Let multiply equation with 2)

solution