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Geometry Difficulty 3.6 AMC 10/12 Find the answer Canada

Points $B$ and $D$ lie on sides $AC$ and $CE$ , respectively, of $\triangle ACE$ , as shown.If $\$\angle CAE=\angle CBD=90\degree$ $and$ CB=BD=DE $, the measure of$ $\angle$ ABE$ is

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Solution

In $\triangle CBD$ , $CB=BD$ and so $\angle BCD$ and $\angle BDC$ have equal measures. The sum of the angles in a triangle is 180 $\degree$ , and since $\angle CBD=90\degree$ , then $\angle BCD=\angle BDC=45\degree$ . (We confirm that $90\degree+45\degree+45\degree=180\degree$ .) Since $\angle CDE$ is a straight angle, then $\$\angle BDE+\angle BDC=180\degree$ $or$ $\angle BDE=180\degree-45\degree=135\degree\$.$

In $\triangle BDE$ , $BD=DE$ and so $\angle DBE$ and $\angle DEB$ have equal measures. The sum of the angles in a triangle is 180 $\degree$ , and since $\angle BDE=135\degree$ , then $\angle DBE=\dfrac{180\degree-135\degree}{2}= 22.5\degree$ .

Finally, since $\angle ABC$ is a straight angle, then $\$\angle ABE+\angle DBE+\angle DBC=180\degree$ $or$ $\angle ABE=180\degree-22.5\degree-90\degree=67.5\degree\$.$

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