Similar triangles

taylor_1989_2012 · September 30th, 2012, 03:36 AM

I am having a bit of trouble finding the length of A-D in the triangle that I have attached to the post. Could someone show how to go about this question.

Question: The diagram shows two similar triangles ABC and BCD. BC=7cm, DC=5cm and angle BAD= angle CBD. Calculate length AD.

I feel I should be to Pythagoras but as it is not a right hand triangle I have no idea where to start.

agentredlum · September 30th, 2012, 04:36 AM

If the triangles are simillar then $\frac{AD + 5}{7}= \frac{7}{5}$

$5AD + 25= 49$

$AD= \frac{24}{5}$

There is another possibility, AD = 0 , then ABC is exactly on top of DBC or vice versa.

September 30th, 2012, 04:36 AM	#2
agentredlum Math Team Joined: Jul 2011 From: North America, 42nd parallel Posts: 3,372 Thanks: 233	Re: Similar triangles If the triangles are simillar then $\frac{AD + 5}{7}= \frac{7}{5}$ $5AD + 25= 49$ $AD= \frac{24}{5}$ There is another possibility, AD = 0 , then ABC is exactly on top of DBC or vice versa.
$agentredlum is offline$

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