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

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

Grade:11

2 Answers

Arun
25763 Points
one year 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
one year 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}

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