Draw a triangle ABC in which AB = AC and a circle through B touches AC at D and intersects AB at P. If D is the mid-point of AC, then prove that AB = 4AP.
Answers
Hence it is proved that AB = 4AP.
Steps of construction:
Draw a line BC with known measure/scale.
From point B draw an arc with a scale of more than BC, above the line BC.
Similarly, from point C draw an arc with a scale of more than BC, above the line BC.
A point where these arcs meet mark it as point A.
Join AB and AC.
Mark a point D at the mid of the lines AC.
Take a radius of scale from point D such that, it touches the point B, when drawn a circle.
Mark a point P, where the line AB and the circle drawn meet.
Hence the required and labelled figure.
To prove: AB = 4AP
we have, AD^2 = AP * AB ..........(1) (using tangent - secant theorem)
AB = AC, D mid point of AC. (given)
AD = AC/2 = AB/2
∴ from (1)
(AB/2)^2 = AP * AB
AB^2 / 4 = AP * AB
AB / 4 = AP
∴ AB = 4 AP
Hence proved.
