Question

Plzzzzz answer this question as fast as u can...
I really need this to be done now. ​

Answer 1

Answer:to prove=9 AD,2 =7AB,2cons.= draw AL prpendicular on BCproof-= in Tri..ALB and Tri..ALC ,we have angle ALB =angle ALC. (each. 90) AL=Al. (common)Tri..ALB congruence to Tri..ALC. (RHS criteria) BL= CL. (cpct)DL = BL- BD 1/2BC-1/3BC 3BC-2BC/2= 1/6BC in tri..ALD angle ALD =90 AD,2=AL,2+DL2..... (by Pythagoras) AL,2 =AD,2 DL,2.......(1) in Tri...ALB , angle ALB =90 similarly. AL,2 == AB2- BC,2.......(2) clearly ,eq(1)and (2)are value of AL then they are equal. AD,2 - DL,2 =AB,2- BL,2 AD,2 - 1/36BC,2. =AB,2 =-1/4 BC,2. (DL = 1/6BC and BL=1/2BC)AD,2= AB,2 (( 1-1/4+1/36)). (BC=AB) AD,2= AB,2.28/36 AD,2=7 AB,2