Sample Problem 1. Given that AB = AC and that BD = DC, prove that AD is perpendicular to BC.
First of all, note that proving that AD is perpendicular to BC is the same as proving that \(\measuredangle ADB=\measuredangle ADC\) because those two angles add to 180 degrees. Seeing information regarding two sides of triangles ABD and ACD, we try to find information on the third side of the triangles because both triangles contain one of the angles we want to prove equal. However, we can easily see that the AD is the remaining side of both triangles, and it is obvious that AD = AD. Thus, we note that all three sides of triangles ACD are congruent to the corresponding sides of triangle ABD. From here, we can see that this implies that triangles ACD and ABD are congruent. In other words, the triangles are essentially the same and can be one can me moved onto the other with only translations, rotations, and reflections and all corresponding sides and angles are equal. Note that when stating a congruence, the order of the points in each shape matter because they are used to show how the two shapes map to each other (the first points named by the first letter in the name of each shape map to each other, and so on). Thus, we see that \(\measuredangle ADB=\measuredangle ADC\) as desired, and we are done.
The claim that these two triangles are congruent may not seem obvious at first. Let's go into more detail to see how we could prove that. Here, we will prove it by showing that given three side lengths, any triangle constructed with those side lengths must be congruent. To do this we will start with translate and rotate any triangle satisfying these conditions such that the side with the longest length is mapped onto AB. From here, we will look at all the possible triangles that could have been created. If a triangle does have the above three lengths and longest sides at AB, then we have two possibilities in deciding the length of the side that comes out of A. From there, we draw a circle from A with a radius equal to the length of the side coming out of A and do the same for point B. We thus see that the third point of our triangle is the point on the intersection between these two circles, leaving two possibilities for where the point is. However, because we could switch the lengths of the sides coming out of A and B, we see that there are actually 4 possible locations for the third point of our triangle.
The claim that these two triangles are congruent may not seem obvious at first. Let's go into more detail to see how we could prove that. Here, we will prove it by showing that given three side lengths, any triangle constructed with those side lengths must be congruent. To do this we will start with translate and rotate any triangle satisfying these conditions such that the side with the longest length is mapped onto AB. From here, we will look at all the possible triangles that could have been created. If a triangle does have the above three lengths and longest sides at AB, then we have two possibilities in deciding the length of the side that comes out of A. From there, we draw a circle from A with a radius equal to the length of the side coming out of A and do the same for point B. We thus see that the third point of our triangle is the point on the intersection between these two circles, leaving two possibilities for where the point is. However, because we could switch the lengths of the sides coming out of A and B, we see that there are actually 4 possible locations for the third point of our triangle.
Finally, as seen in the diagram above, once we have one of the triangles corresponding to a possible location of the third point, we can reflect it over AB, the perpendicular bisector of AB, and both lines, in order to produce three more, for a total of 4, congruent triangles corresponding to a possible location of the third point. However, because there are only 4 possible locations for the third point, as described above, we have just covered each of these possible locations and showed that each creates a triangle that can be mapped on to each other triangle with only reflections. Thus, we have proved that any triangle with three side lengths is congruent, and we are done. If you don't understand this proof, go over it again until you do, because it is very important.
Now that you've seen that we can prove two triangles congruent without explicitly proving all of their sides and all of their angles congruent, you may wonder if there are other sets of givens that can prove two shapes congruent. Indeed, two triangles can be proved congruent in several ways, which will be listed below. For each such method of proving two shapes congruent, try to convince yourself that any two triangles sharing these characteristics must indeed be congruent. Finally, try to remember the list given below. Congruent triangles are the key to many geometry problems, so you will need to know what to look for to prove two triangles congruent.
Possible ways to prove triangles congruent
Now that you've seen that we can prove two triangles congruent without explicitly proving all of their sides and all of their angles congruent, you may wonder if there are other sets of givens that can prove two shapes congruent. Indeed, two triangles can be proved congruent in several ways, which will be listed below. For each such method of proving two shapes congruent, try to convince yourself that any two triangles sharing these characteristics must indeed be congruent. Finally, try to remember the list given below. Congruent triangles are the key to many geometry problems, so you will need to know what to look for to prove two triangles congruent.
Possible ways to prove triangles congruent
- Two triangles are congruent if all 3 pairs of sides are congruent (called SSS congruence).
- Two triangles are congruent if they have two pairs of congruent sides and the angle between those two sides is also congruent in both triangles (called SAS congruence).
- Two triangles are congruent if they have a right angle and their hypotenuses as well as one pair of corresponding legs are congruent (called HL congruence).
- Two triangles are congruent if two angles and the side between them are congruent (called ASA congruence).
Sample Problem 2. Using the given lengths in the diagram below, find AD.
For this problem it isn't immediately obvious what to do, so we will start by seeing if we can find any additional lengths using the given information. Because ABA' is a right triangles, we can use the Pythagorean theorem to see that AA' = 5. Seeing that we now have a lot of information about lengths for triangles ABA', we will try to see if we can relate this triangle to some triangle that contains AD in order to try to find the length of AD. Notice that because angle A is congruent to itself, \(\measuredangle ACD=\measuredangle ABA'=90^{\circ}\), and the angles of a triangle add to 180 degrees, the angles of triangle ACD are all congruent to corresponding angles of triangle ABA'. Thus, these two triangles seem highly related, but they definitely aren't congruent. These triangles are similar triangles, which means that you can move one on to the other using translations, rotations, reflections, and scaling/dilation. We can see that, by changing the length of each side of triangle ACD by moving points C and D along sides AB and AA' respectively, we can essentially map triangle ACD onto triangle ABA' because CD is parallel to BA'. This transformation is called a dilation. Because CD maps to BA', we can see that this dilation mapping ACD to ABA' doubles the lengths of each side of triangle ACD. You can see that AC and AD must be scaled the same amount because CD is parallel to A'B, so, using the Pythagorean theorem, we see that doubling A'B requires doubling the other lengths. In order to generalize this for other problems that aren't just right triangles being dilated, think of splitting triangles into two right triangles, or expressing an acute triangle as one right triangle minus another, as seen in the diagram below. Thus, we can conclude AD is half of AA' and has a length of 2.5. Make sure you understand this solution, because it is very important.
There are several ways to prove two triangles similar. For similar triangles ABC and DEF, it can be shown that AB/DE=BC/EF=AC/DF and that angles A and D, B and E, and C and F are all congruent. Knowing that two triangles are similar can be used to find lengths or angles, as in the problem above. Thus, it is important to know the criteria for two triangles to be similar. A list of ways to show two triangles similar is given below, and for each way you should try to convince yourself that the given conditions do indeed make two triangles similar.
Ways to show two triangles similar
Ways to show two triangles similar
- Two triangles are similar if all three of their corresponding pairs of angles are congruent (two pairs of corresponding angles being congruent is enough because all angles of a triangle add to 180 degrees). This is called AA similarity.
- If three pairs of corresponding sides all have the same ratio between the two triangles, then the triangles are similar (called SSS similarity).
- If two pairs of corresponding sides have the same ratio between triangles and the angle between them is the same for each triangle, then the two triangles are similar (called SAS similarity).
Exercises
- Prove that the ratio between each pair of corresponding sides is the same for two triangles who have all of their corresponding angles congruent.
- Find EF in terms of FB and FC.
Solution 1
We just about proved this problem already in sample problem two. All we need to do for this, is rotate and translate the smaller of the two similar triangles so that one of its angles coincides with an angle of the larger triangle, such that the two sides around the angle in each triangle are co-linear. The two triangles will then look like one of the three diagrams in sample problem 2, and we can use the Pythagorean theorem as described in sample problem 2 to finish the problem off.
SOlution 2
Note that triangle EFB is similar to triangle CEB because angle EFB equals angle CEB ancf/d angle B equals itself. Thus, we get that EF/FB=CE/EB. Similarly, we see that triangle EFC is similar to triangle CEB, meaning that CF/EF=CE/EB. Thus, we have EF/FB=CF/EF or, equivalently, \(EF^2=FB\cdot CF\implies EF=\sqrt{FB\cdot CF}\).