Question

the base bc of an isosceles triangle abc is produced to d and a straight line pq is drawn cutting ac internally at p and meeting ab in q prove that ap is greater than aq

Answer 1

In isosceles  Δ a b c

a b = a c  (given)

A base b c is produced to point d , and a line is drawn from point d such that it cuts side a c and a b at p and q respectively.

Since it is not perpendicular distance , the line which is nearer have the smaller segment .

So, q b > p c

- q b < - p c

a b - q b< a b - p c  →→→subtracting from a b on both sides of inequality

a q < a c - p c

a q < a p

Hence proved.

Answer 2

Answer:

Step-by-step explanation:

In isosceles  Δ a b c

a b = a c  (given)

A base b c is produced to point d , and a line is drawn from point d such that it cuts side a c and a b at p and q respectively.

Since it is not perpendicular distance , the line which is nearer have the smaller segment .

So, q b > p c

- q b < - p c

a b - q b< a b - p c  →→→subtracting from a b on both sides of inequality

a q < a c - p c

a q < a p

Hence proved.