Question

D is the mid point of bc and ae perpendicular bc.ifBC=aAC=b,AB=c,ED=x,AD=p and AE=h,prove that:b2=p2+ax+a2÷4

Answer 1

This is the answer hope it helps

Answer 2

Answer:

Step-by-step explanation:

(i) In right-angled triangle AEC, applying Pythagoras theorem, we have:

AC²=AE²+EC²

=> b²=h²+(x+a/2)²

=>b²=h²+x²+a²/4+ax.....(i)

In right angle triangle AED we get

AD²=AE²+ED²

=>p²=h²+x² …(ii)

Therefore from (i) and (ii)

b²=p²+ax+a²/x

(ii) In right-angled triangle AEB, applying Pythagoras, we have:

AB²=AE²+EB²

=> c²=h²+(a/2–x)²(As BD=a/2 and BE=BD–x)

=> c²=h²+x²–a²/4(h²+x²=p²)

=> c²=p²–ax+a²/4

(iii) Adding (i) and (ii), we get:

=> b²+c²= p²+ax+a²/4+p²–ax+a²/4

= 2p²+ax–ax+(a²+a²)/4

= 2p²+a²/2