Question

ABC is an acute angled triangle from A perpendicular AD is drawn on BC prove that AC^2=AB^2+ BC^2-2BC*BD​

Answer 1

Step-by-step explanation:

In △ABC,∠ABD=90°

By pathagores theorem

AB

2

=AD

2

+BD

2

In △ADC,∠ADC=90°

AC

2

=AD

2

+DC

2

AC

2

=AD

2

+(BC−BD)

2

AC

2

=AD

2

+BC

2

+BD

2

−BC×BD

AC

2

=(AD

2

+BD

2

)+BC

2

−BC×BD

AC

2

=AB

2

+BC

2

−BC×BD

hence proved