# Consider a parallelogram ABCD. Suppose that P is a point onthe side AD so that AP : AD = x : y. Let Q be the intersectionpoint of AC and PB. Show that AQ : AC = x : x + y

Arun
25758 Points
2 years ago
Dear student

There is no image attached. Please check and repost the question with a proper attachment. I will be happy to help you.

Thanks and regards
Shivani
14 Points
2 years ago
Hello,
I am providing solution without diagram as I am not able to upload it. You can consider the diagram as per the question.

In triangle QPA & triangle QBC,
$\angle QPA = \angle QBC$           ….... alternate angles
$\angle PQA = \angle BQC$           ….... vertically opposite angles
therefore triangle QPA is similar with triangle QBC     …....by AA test of similarity
$\therefore \frac{BC}{AP} = \frac{CQ}{AQ}$
but AD = BC = y   …...... opposite sides of parallelogram
$\therefore \frac{CQ}{AQ} = \frac{y}{x}$
$\therefore \frac{CQ+AQ}{AQ} = \frac{y+x}{x}$
$\therefore \frac{AC}{AQ} = \frac{x+y}{x}$
$\therefore \frac{AQ}{AC} = \frac{x}{x+y}$