Geometry Proof, Need help!!!

[lynlawdac]

Hi,

I have the following problem that I am trying to explain to my son and I don't know where to start or finish for that matter. Here it is:

Given: Triangle PDQ with line segment AB parallel to line segment PQ

Prove: DA x PQ = AB x DP

I need the step by step proof with the statements and reasons.


Thanks to anyone who could help me out on this one.

 

[tkhunny]

You didn't state where A and B are? On the triangle maybe? If so, where? Does it matter?

 

[pka]

Given: Triangle PDQ with line segment AB parallel to line segment PQ
Prove: DA x PQ = AB x DP

First you must add that: A is between P & D.
Then prove ΔDAB similar to ΔDPQ.
Do that by corresponding angles from parallel lines.
Then you have DA/DP = AB/PQ.

 

[lynlawdac]

Steps for proof?

Can anyone give me a step by step proof for this one? Maybe it would make more sense to me if I saw how it plays out.

 

[pka]

Because \(\displaystyle \overline {AB} ||\overline {PQ}\) then \(\displaystyle \angle DAB \simeq \angle DPQ\quad \& \quad \angle DBA \simeq \angle DQP\).

This is sufficient to show \(\displaystyle \Delta DAB \approx \Delta DPQ\).

By corresponding parts we get \(\displaystyle \frac{{DA}}{{DP}} = \frac{{AB}}{{PQ}}\).