Question

answer this question ​

Answer 1

its very simple dude

it can be solved by two method,first one is using similarty property and other by using mid- point theorem, which u had read in class 9th

let solve it by mid point theorem

mid- point theorem ,states that if a line intersect the two sides of triangle,and parallel to the third line,then it will be 1/2,to the opposite line

so here

in triangle APC,angleQBC=anglePAB

so PAis parellel to QB

also QB=1/2 PA. { by mid-point theorem}

so,y=x/2

1/y=2/x

1/x=1/2y.........1)

similarly in triangle RAC,,QB is parallel to RC

so,QB=RC/2

z=2y.........(by mid- point theorem)

1/z=1/2y......2)

add both equation

1/x +1/z=1/2y +1/2y

1/x+1/z=1/y

hope it helps you

Answer 2

Solution :-

Consider ΔPAC and ΔQBC

PA ⊥ AC, QB⊥ AC

⇒ ∠PAC = ∠QBC = 90°

∠ACP is common

Third angles are also equal

Therefore by AAA similarity ΔPAC ~ ΔQBC

⇒ QB/PA = BC/AC [ Corresponding sides of similar triangles ]

⇒ y/x = BC/AC --> EQ(1)

Consider ΔRCA and ΔQBA

RC ⊥ AC, QB ⊥ AC

⇒ ∠RCA =∠QBA = 90°

∠RAC is common

Third angles are also equal

Therefore by AAA similarity ΔRCA ~ ΔQBA

⇒ QB/RC = AB/AC [ Corresponding sides of similar triangles ]

⇒ y/z = AB/AC ---> eq(2)

Adding eq(1) and eq(2)

⇒ y/x + y/z = BC/AC + AB/AC

⇒ y(1/x + 1/z) = (BC + AB) /AC

⇒ y(1/x + 1/z) = AC/AC [ From figure, BC + AB = AC ]

⇒ y(1/x + 1/z) = 1

⇒ 1/x + 1/z = 1/y

Hence proved.