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Grade 11,

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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
25750 Points
3 years ago
Dear student
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There is no image attached. Please check and repost the question with a proper attachment. I will be happy to help you.
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Thanks and regards
Shivani
14 Points
3 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.
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In triangle QPA & triangle QBC,
\angle QPA = \angle QBCย  ย  ย  ย  ย  ย โ€ฆ.... alternate angles
\angle PQA = \angle BQCย  ย  ย  ย  ย  ย โ€ฆ.... vertically opposite angles
therefore triangle QPA is similar with triangle QBCย  ย  ย โ€ฆ....by AA test of similarity
\therefore \frac{BC}{AP} = \frac{CQ}{AQ}
but AD = BC = yย  ย โ€ฆ...... 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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