# Finding the sides with equal angles

• September 7th 2009, 08:55 AM
saberteeth
Finding the sides with equal angles
In the given figure BA is parallel to CD, $\angle DAC =\angle ABC$. AB = 10cm, BC= 9cm, AC=15cm. What is the length of AD?
I tried solving with AB/BC = AD/AC. The answer turned out to be wrong because the solution in the book is given as,
Since $\triangle$ ABC is similar to $\triangle$CAD, therefore, AB/BC = AC/AD.

I did not understand why they considered $\triangle$CAD instead $\triangle$DAC and AB/BC = AC/AD accordingly.
• September 7th 2009, 09:18 AM
Plato
Quote:

Originally Posted by saberteeth
I did not understand why they considered $\triangle$CAD instead $\triangle$DAC and AB/BC = AC/AD accordingly.

In all cases involving similarly or congruence, the order in which we list the vertices in all important.
In this case it must be $\Delta ABC \approx \Delta CAD$ because $\angle ABC \cong \angle CAD\;\& \,\angle BAC \cong \angle ACD$.
• September 7th 2009, 09:26 AM
saberteeth
Quote:

Originally Posted by Plato
In all cases involving similarly or congruence, the order in which we list the vertices in all important.
In this case it must be $\Delta ABC \approx \Delta CAD$ because $\angle ABC \cong \angle CAD\;\& \,\angle BAC \cong \angle ACD$.

Oh thanks, now i understand..