Question

In ΔABC , ∠B=900 and seg BD⊥ side AC, A-D-C,

then ______​

Answer 1

seriously broo i cant understand what r u try to saying ... i posted the answer that i think will help u if not pls forgive me..

ABC is a right angled triangle in which angle A=90° and AD is perpendicular

to BC. Thus , In right angled triangle ABC , AB^2+AC^2=BC^2……………(1)

In right angled triangle ADB

AD^2+DB^2=AB^2……………….(2)

In right angled triangle ADC

AD^2+DC^2=AC^2……………….(3)

by adding the eqn. (2) and (3)

2.AD^2+DB^2+DC^2= AB^2+AC^2. , putting AB^2+AC^2=BC^2 from eqn. (1)

2.AD^2+DB^2+DC^2=BC^2. , putting BC=BD+CD

2.AD^2+DB^2+DC^2=(BD+CD)^2

or. 2.AD^2+DB^2+DC^2=BD^2+CD^2+2×BD×CD.

or. 2.AD^2 = 2×BD×CD.