A point has no dimension. It is usually represented by a small dot. A PointA U SING U NDEFINED T ERMS AND D EFINITIONS

A line extends in one dimension. It is usually represented by a straight line with two arrowheads to indicate that the line extends without end in two directions. Lineor AB Collinear points are points that lie on the same line. U SING U NDEFINED T ERMS AND D EFINITIONS

A plane extends in two dimensions. It is usually represented by a shape that looks like a table or a wall, however you must imagine that the plane extends without end. U SING U NDEFINED T ERMS AND D EFINITIONS

Coplanar points are points that lie on the same plane. Plane M or plane ABC A C M B U SING U NDEFINED T ERMS AND D EFINITIONS

Naming Collinear and Coplanar Points S OLUTION Points D, E, F lie on the same line, so they are collinear. There are many correct answers. For instance, points H, E, and G do not lie on the same line. Points D, E, F, and G lie on the same plane, so they are coplanar. Also, D, E, F, and H are coplanar. G H F E D Name three points that are collinear. Name four points that are coplanar. Name three points that are not collinear.

Another undefined concept in geometry is the idea that a point on a line is between two other points on the line. You can use this idea to define other important terms in geometry. U SING U NDEFINED TERMS AND DEFINITIONS

The line segment or segment AB (symbolized by AB) consists of the endpoints A and B, and all points on AB that are between A and B. U SING U NDEFINED T ERMS AND D EFINITIONS Consider the line AB (symbolized by AB).

The ray AB (symbolized by AB) consists of the initial point A and all points on AB that lie on the same side of A as point B. Note that AB is the same as BA, and AB is the same as BA. However, AB and BA are not the same. They have different initial points and extend in different directions. U SING U NDEFINED T ERMS AND D EFINITIONS

If C is between A and B, then CA and CB are opposite rays. Like points, segments and rays are collinear if they lie on the same plane. So, any two opposite rays are collinear. Segments, rays, and lines are coplanar if they lie on the same plane. U SING U NDEFINED T ERMS AND D EFINITIONS

Drawing Lines, Segments, and Rays Draw three noncollinear points J, K, L. Then draw JK, KL and L J. S OLUTION Draw J, K, and L Draw JK. Draw LJ. K L J Draw KL.

Drawing Opposite Rays Draw two lines. Label points on the lines and name two pairs of opposite rays. S OLUTION Points M, N, and X are collinear and X is between M and N. So, XM and XN are opposite rays. Points P, Q, and X are collinear and X is between P and Q. So, XP and XQ are opposite rays.

S KETCHING I NTERSECTIONS OF L INES AND P LANES Two or more geometric figures intersect if they have one or more points in common. The intersection of the figures is the set of points the figures have in common.

Sketching Intersections Sketch a line that intersects a plane at one point. S OLUTION Draw a plane and a line. Emphasize the point where they meet. Dashes indicate where the line is hidden by the plane.

Sketch two planes that intersect in a line. Sketching Intersections S OLUTION Draw two planes. Emphasize the line where they meet. Dashes indicate where one plane is hidden by the other plane

C ONGRUENCE OF A NGLES THEOREM THEOREM 2.2 Properties of Angle Congruence Angle congruence is r ef lex ive, sy mme tric, and transitive. Here are some examples. TRANSITIVE IfA  BandB  C, then A  C SYMMETRIC If A  B, then B  A REFLEX IVE For any angle A, A  A

Transitive Property of Angle Congruence Prove the Transitive Property of Congruence for angles. S OLUTION To prove the Transitive Property of Congruence for angles, begin by drawing three congruent angles. Label the vertices as A, B, and C. GIVEN A B, PROVE A CA C A B C B CB C

Transitive Property of Angle Congruence GIVEN A B, B CB C PROVE A CA C StatementsReasons mA = mB Definition of congruent angles 5 A  C Definition of congruent angles A  B,Given B  C mB = mC Definition of congruent angles mA = mC Transitive property of equality

Using the Transitive Property This two-column proof uses the Transitive Property. StatementsReasons m1 = m3 Definition of congruent angles GIVEN m3 = 40°,12,23 PROVE m1 = 40° 1 m1 = 40° Substitution property of equality 13 Transitive property of Congruence Givenm3 = 40°,12, 2323

Proving Theorem 2.3 THEOREM THEOREM 2.3 Right Angle Congruence Theorem All right angles are congruent. You can prove Theorem 2.3 as shown. GIVEN 1 and2 are right angles PROVE 1212

Proving Theorem 2.3 StatementsReasons m1 = 90°, m2 = 90° Definition of right angles m1 = m2 Transitive property of equality 1  2 Definition of congruent angles GIVEN 1 and2 are right angles PROVE and2 are right angles Given

P ROPERTIES OF S PECIAL P AIRS OF A NGLES THEOREMS THEOREM 2.4 Congruent Supplements Theorem If two angles are supplementary to the same angle (or to congruent angles) then they are congruent

P ROPERTIES OF S PECIAL P AIRS OF A NGLES THEOREMS THEOREM 2.4 Congruent Supplements Theorem If two angles are supplementary to the same angle (or to congruent angles) then they are congruent If m1 + m2 = 180° m2 + m3 = 180° and 1 then 1  3

P ROPERTIES OF S PECIAL P AIRS OF A NGLES THEOREMS THEOREM 2.5 Congruent Complements Theorem If two angles are complementary to the same angle (or to congruent angles) then the two angles are congruent

P ROPERTIES OF S PECIAL P AIRS OF A NGLES THEOREMS THEOREM 2.5 Congruent Complements Theorem If two angles are complementary to the same angle (or to congruent angles) then the two angles are congruent. 4 If m4 + m5 = 90° m5 + m6 = 90° and then 4 

Proving Theorem 2.4 StatementsReasons 1 2 GIVEN 1 and2 are supplements PROVE and4 are supplements and2 are supplementsGiven 3 and4 are supplements 1  4 m1 + m2 = 180° Definition of supplementary angles m3 + m4 = 180°

Proving Theorem 2.4 StatementsReasons 3 GIVEN 1 and2 are supplements PROVE and4 are supplements m1 + m2 = Substitution property of equality m3 + m1 m1 + m2 = Transitive property of equality m3 + m4 m1 = m4 Definition of congruent angles

Proving Theorem 2.4 StatementsReasons GIVEN 1 and2 are supplements PROVE and4 are supplements m2 = m3 Subtraction property of equality 23 Definition of congruent angles

POSTULATE POSTULATE 12 Linear Pair Postulate If two angles for m a linear pair, then they are supplementary. m1 + m2 = 180° P ROPERTIES OF S PECIAL P AIRS OF A NGLES

Proving Theorem 2.6 THEOREM THEOREM 2.6 Vertical Angles Theorem Vertical angles are congruent 1 3,24

Proving Theorem 2.6 PROVE 5757 GIVEN 5 and6 are a linear pair, 6 and7 are a linear pair StatementsReasons 5 and6 are a linear pair, Given 6 and7 are a linear pair 5 and6 are supplementary, Linear Pair Postulate 6 and7 are supplementary 5 7 Congruent Supplements Theorem

P ROPERTIES OF P ARALLEL L INES POSTULATE POSTULATE 15 Corresponding Angles Postulate If two parallel lines are cut by a transversal, then the pairs of corresponding angles are congruent.

P ROPERTIES OF P ARALLEL L INES THEOREMS ABOUT PARALLEL LINES THEOREM 3.4 Alternate Interior Angles If two parallel lines are cut by a transversal, then the pairs of alternate interior angles are congruent.

P ROPERTIES OF P ARALLEL L INES THEOREMS ABOUT PARALLEL LINES THEOREM 3.5 Consecutive Interior Angles 5 6 m 5 + m 6 = 180° If two parallel lines are cut by a transversal, then the pairs of consecutive interior angles are supplementary.

P ROPERTIES OF P ARALLEL L INES THEOREMS ABOUT PARALLEL LINES THEOREM 3.6 Alternate Exterior Angles If two parallel lines are cut by a transversal, then the pairs of alternate exterior angles are congruent.

P ROPERTIES OF P ARALLEL L INES THEOREMS ABOUT PARALLEL LINES THEOREM 3.7 Perpendicular Transversal j k If a transversal is perpendicular to one of two parallel lines, then it is perpendicular to the other.

Proving the Alternate Interior Angles Theorem Prove the Alternate Interior Angles Theorem. S OLUTION GIVEN p || q p || qGiven StatementsReasons PROVE  3 Corresponding Angles Postulate 3  2 Vertical Angles Theorem 1  2 Transitive property of Congruence

Using Properties of Parallel Lines S OLUTION Given that m 5 = 65°, find each measure. Tell which postulate or theorem you use. Linear Pair Postulate m 7 = 180° – m 5 = 115° Alternate Exterior Angles Theorem m 9 = m 7 = 115° Corresponding Angles Postulate m 8 = m 5 = 65° m 6 = m 5 = 65° Vertical Angles Theorem

Using Properties of Parallel Lines Use properties of parallel lines to find the value of x. S OLUTION Corresponding Angles Postulate m 4 = 125° Linear Pair Postulate m 4 + (x + 15)° = 180° Substitute. 125° + (x + 15)° = 180° P ROPERTIES OF S PECIAL P AIRS OF A NGLES Subtract. x = 40°

Over 2000 years ago Eratosthenes estimated Earth’s circumference by using the fact that the Sun’s rays are parallel. When the Sun shone exactly down a vertical well in Syene, he measured the angle the Sun’s rays made with a vertical stick in Alexandria. He discovered that Estimating Earth’s Circumference: History Connection m of a circle

Estimating Earth’s Circumference: History Connection m of a circle Using properties of parallel lines, he knew that m 1 = m 2 He reasoned that m of a circle

The distance from Syene to Alexandria was believed to be 575 miles Estimating Earth’s Circumference: History Connection m of a circle Earth’s circumference 1 50 of a circle 575 miles Earth’s circumference 50(575 miles) Use cross product property 29,000 miles How did Eratosthenes know that m 1 = m 2 ?

Estimating Earth’s Circumference: History Connection How did Eratosthenes know that m 1 = m 2 ? S OLUTION Angles 1 and 2 are alternate interior angles, so 1  2 By the definition of congruent angles, m 1 = m 2 Because the Sun’s rays are parallel,

SSS AND SAS C ONGRUENCE P OSTULATES If all six pairs of corresponding parts (sides and angles) are congruent, then the triangles are congruent. and then If Sides are congruent 1. AB DE 2. BC EF 3. AC DF Angles are congruent 4. AD 5. BE 6. CF Triangles are congruent  ABC  DEF

SSS AND SAS C ONGRUENCE P OSTULATES POSTULATE POSTULATE 19 Side - Side - Side (SSS) Congruence Postulate Side MNQR Side PMSQ Side NPRS If If three sides of one triangle are congruent to three sides of a second triangle, then the two triangles are congruent. then  MNP  QRS SSS

SSS AND SAS C ONGRUENCE P OSTULATES The SSS Congruence Postulate is a shortcut for proving two triangles are congruent without using all six pairs of corresponding parts.

Using the SSS Congruence Postulate Prove that  PQW  TSW. Paragraph Proof S OLUTION So by the SSS Congruence Postulate, you know that  PQW   TSW. The marks on the diagram show that PQ  TS, PW  TW, and QW  SW.

POSTULATE SSS AND SAS C ONGRUENCE P OSTULATES POSTULATE 20 Side-Angle-Side (SAS) Congruence Postulate Side PQWX Side QSXY then  PQS  WXY Angle QX If If two sides and the included angle of one triangle are congruent to two sides and the included angle of a second triangle, then the two triangles are congruent. ASS

1 Using the SAS Congruence Postulate Prove that  AEB  DEC. 2 3  AEB   DEC SAS Congruence Postulate 21 AE  DE, BE  CEGiven 1  2Vertical Angles Theorem StatementsReasons

D G A R Proving Triangles Congruent M ODELING A R EAL- L IFE S ITUATION PROVE  DRA  DRG S OLUTION A RCHITECTURE You are designing the window shown in the drawing. You want to make  DRA congruent to  DRG. You design the window so that DR AG and RA  RG. Can you conclude that  DRA   DRG ? GIVEN DR AG RA RG

SAS Congruence Postulate  DRA   DRG 1 Proving Triangles Congruent Given DR AG If 2 lines are, then they form 4 right angles. DRA and DRG are right angles. Right Angle Congruence Theorem DRA  DRG Given RA  RG Reflexive Property of Congruence DR  DR StatementsReasons D GAR GIVEN PROVE  DRA  DRG DR AG RA RG

Congruent Triangles in a Coordinate Plane AC  FH AB  FG AB = 5 and FG = 5 S OLUTION Use the SSS Congruence Postulate to show that  ABC   FGH. AC = 3 and FH = 3

Congruent Triangles in a Coordinate Plane d = (x 2 – x 1 ) 2 + ( y 2 – y 1 ) 2 = = 34 BC = (– 4 – (– 7)) 2 + (5 – 0 ) 2 d = (x 2 – x 1 ) 2 + ( y 2 – y 1 ) 2 = = 34 GH = (6 – 1) 2 + (5 – 2 ) 2 Use the distance formula to find lengths BC and GH.

Congruent Triangles in a Coordinate Plane BC  GH All three pairs of corresponding sides are congruent,  ABC   FGH by the SSS Congruence Postulate. BC = 34 and GH = 34

U SING M EDIANS OF A T RIANGLE A median of a triangle is a segment whose endpoints are a vertex of the triangle and the midpoint of the opposite side. D

U SING M EDIANS OF A T RIANGLE The three medians of a triangle are concurrent. The point of concurrency is called the centroid of the triangle. The centroid is always inside the triangle. acute triangleright triangleobtuse triangle centroid The medians of a triangle have a special concurrency property.

THEOREM THEOREM 5.7 Concurrency of Medians of a Triangle U SING M EDIANS OF A T RIANGLE The medians of a triangle intersect at a point that is two-thirds of the distance from each vertex to the midpoint of the opposite side. If P is the centroid of  ABC, then BP = BF 2 3 P AP = AD 2 3 CP = CE 2 3

U SING M EDIANS OF A T RIANGLE The centroid of a triangle can be used as its balancing point. A triangular model of uniform thickness and density will balance at the centroid of the triangle.

Using the Centroid of a Triangle P is the centroid of  QRS shown below and PT = 5. Find RT and RP. S OLUTION So, RP = 10 and RT = 15. Then RP = RT = (15) = Because P is the centroid, RP = RT. 2 3 Substituting 5 for PT, 5 = RT, so RT = Then PT = RT – RP = RT 1 3

Finding the Centroid of a Triangle Find the coordinates of the centroid of  JKL. S OLUTION The centroid is two thirds of the distance from each vertex to the midpoint of the opposite side. (5, 8) Choose the median KN. Find the coordinates of N, the midpoint of JL. (7, 10) (3, 6) (5, 2) , = 10 2, 16 2 = (5, 8) The coordinates of N are J K L N P

Find the coordinates of the centroid of  JKL. S OLUTION J K L N (5, 8) (7, 10) (3, 6) (5, 2) P M Find the distance from vertex K to midpoint N. The distance from K (5,2) to N(5,8) is 8 – 2, or 6 units. [Yellow coordinates appear.] Determine the coordinates of the centroid, which is 6, or 4 units up from vertex K along the median KN. 2 3 Finding the Centroid of a Triangle The coordinates of the centroid P are (5, 2 + 4), or (5, 6) (5, 6)

An altitude of a triangle is the perpendicular segment from a vertex to the opposite side or to the line that contains the opposite side. An altitude can lie inside, on, or outside the triangle. Every triangle has three altitudes. The lines containing the altitudes are concurrent and intersect at a point called the ortho center of the triangle. U SING A LTITUDES OF A T RIANGLE

Drawing Altitudes and Orthocenters Where is the ortho center of an acute triangle? S OLUTION Draw an example. The three altitudes intersect at G, a point inside the triangle.

Drawing Altitudes and Orthocenters The two legs, LM and KM, are also altitudes. They intersect at the triangle’s right angle. This implies that the orthocenter is on the triangle at M, the vertex of the right angle of the triangle. Where is the orthocenter of a right triangle? S OLUTION

Drawing Altitudes and Orthocenters The three lines that contain the altitudes intersect at W, a point outside the triangle. Where is the orthocenter of an obtuse triangle? S OLUTION

THEOREM U SING A LTITUDES OF A T RIANGLE THEOREM 5.8 Concurrency of Altitudes of a Triangle The lines containing the altitudes of a triangle are concurrent. If AE, BF, and CD are the altitudes of  ABC, then the lines AE, BF, and CD intersect at some point H.

P ROPERTIES OF S PEC IAL P ARALLELOGRAMS | | | | A rectangle is a parallelogram with four right angles. | | | | A square is a parallelogram with four congruent sides and four right angles. A rhombus is a parallelogram with four congruent sides.

P ROPERTIES OF S PEC IAL P ARALLELOGRAMS The Venn diagram shows the relationships among parallelograms, rhombuses, rectangles, and squares. Each shape has the properties of every group that it belongs to. For instance, a square is a rectangle, a rhombus, and a parallelogram, so it has all of the properties of each of those shapes.

P ROPERTIES OF S PEC IAL P ARALLELOGRAMS parallelograms rhombusesrectangles squares

S OLUTION Describing a Special Parallelogram Decide whether the statement is always, sometimes, or never true. A rhombus is a rectangle. The statement is sometimes true. In the Venn Diagram, the regions for rhombuses and rectangles overlap. If the rhombus is a square, it is a rectangle. Help

S OLUTION The statement is sometimes true. Some parallelograms are rectangles. In the Venn diagram, you can see that some of the shapes in the parallelogram box are in the region for rectangles, but many aren’t. Describing a Special Parallelogram A parallelogram is a rectangle. Help Decide whether the statement is always, sometimes, or never true.

Using Properties of Special Parallelograms ABCD is a rectangle. What else do you know about ABCD ? AB DC S OLUTION Because ABCD is a rectangle, it has four right angles by the definition. The definition also states that rectangles are parallelograms, so ABCD has all the properties of a parallelogram: 1 Opposite sides are parallel and congruent. 2 Opposite angles are congruent and consecutive angles are supplementary. 3 Diagonals bisect each other.

Using Properties of Special Parallelograms ABCD is a rectangle. What else do you know about ABCD ? AB DC A rectangle is defined as a parallelogram with four right angles. But any quadrilateral with four right angles is a rectangle because any quadrilateral with four right angles is a parallelogram.

Using Properties of Special Parallelograms COROLLARIES ABOUT SPECIAL QUADRILATERALS RHOMBUS COROLLARY A quadrilateral is a rhombus if and only if it has four congruent sides. RECTANGLE COROLLARY A quadrilateral is a rectangle if and only if it has four right angles. SQUARE COROLLARY A quadrilateral is a square if and only if it is a rhombus and a rectangle. You can use these corollaries to prove that a quadrilateral is a rhombus, rectangle, or square without proving first that the quadrilateral is a parallelogram.

Using Properties of a Rhombus In the diagram, PQRS is a rhombus. What is the value of y? S OLUTION All four sides of a rhombus are congruent, so RS = PS. 5 y – 6 = 2 y + 3 Add 6 to each side. 5 y = 2 y + 9 Subtract 2y from each side. 3 y = 9 Divide each side by 3. y = 3 2y + 3 5y – 6 PQ RS Equate lengths of congruent sides.

THEOREMS U SING D IAGONALS OF S PECIAL P ARALLELOGRAMS THEOREM 6.11 A parallelogram is a rhombus if and only if its diagonals are perpendicular. ABCD is a rhombus if and only if AC BD

THEOREMS U SING D IAGONALS OF S PECIAL P ARALLELOGRAMS THEOREM 6.12 A parallelogram is a rhombus if and only if each diagonal bisects a pair of opposite angles. BC DA ABCD is a rhombus if and only if AC bisects DAB and BCD and BD bisects ADC and CBA

U SING D IAGONALS OF S PECIAL P ARALLELOGRAMS THEOREM S THEOREM 6.13 A parallelogram is a rectangle if and only if its diagonals are congruent. AB DC ABCD is a rectangle if and only if AC  BD

U SING D IAGONALS OF S PECIAL P ARALLELOGRAMS You can rewrite Theorem 6.11 as a conditional statement and its converse. Conditional statement: If the diagonals of a parallelogram are perpendicular, then the parallelogram is a rhombus. Converse: If a parallelogram is a rhombus, then its diagonals are perpendicular. To prove the theorem, you must prove both statements.

P ROPERTIES OF S PEC IAL P ARALLELOGRAMS parallelograms rhombusesrectangles squares Back

U SING S IMILARITY T HEOREMS THEOREM S THEOREM 8.2 Side-Side-Side (SSS) Similarity Theorem If the corresponding sides of two triangles are proportional, then the triangles are similar. If = = A B PQ BC QR CA RP then  ABC ~  PQR. A BC P Q R

U SING S IMILARITY T HEOREMS THEOREM S THEOREM 8.3 Side-Angle-Side (SAS) Similarity Theorem If an angle of one triangle is congruent to an angle of a second triangle and the lengths of the sides including these angles are proportional, then the triangles are similar. then  XYZ ~  MNP. ZX PM XY MN If XM and= X ZY M PN

Proof of Theorem 8.2 GIVEN PROVE = ST MN RS LM TR NL  RST ~  LMN S OLUTION Paragraph Proof M NL RT S PQ Locate P on RS so that PS = LM. Draw PQ so that PQ RT. Then  RST ~  PSQ, by the AA Similarity Postulate, and. = ST SQ RS PS TR QP Use the definition of congruent triangles and the AA Similarity Postulate to conclude that  RST ~  LMN. Because PS = LM, you can substitute in the given proportion and find that SQ = MN and QP = NL. By the SSS Congruence Theorem, it follows that  PSQ   LMN.

E FD A C B GJ H Using the SSS Similarity Theorem Which of the following three triangles are similar? S OLUTION To decide which of the triangles are similar, consider the ratios of the lengths of corresponding sides. Ratios of Side Lengths of  ABC and  DEF = =, 6 4 AB DE Shortest sides = =, 12 8 CA FD Longest sides = BC EF Remaining sides Because all of the ratios are equal,  ABC ~  DEF

= =, CA JG Longest sides E FD Using the SSS Similarity Theorem A C B GJ H Which of the following three triangles are similar? S OLUTION To decide which of the triangles are similar, consider the ratios of the lengths of corresponding sides. Ratios of Side Lengths of  ABC and  GHJ = = 1, AB GH Shortest sides = 9 10 BC HJ Remaining sides Because all of the ratios are not equal,  ABC and  DEF are not similar. E FD A C B GJ H Since  ABC is similar to  DEF and  ABC is not similar to  GHJ,  DEF is not similar to  GHJ.

Using the SAS Similarity Theorem Use the given lengths to prove that  RST ~  PSQ. S OLUTION PROVE  RST ~  PSQ GIVEN SP = 4, PR = 12, SQ = 5, QT = 15 Paragraph Proof Use the SAS Similarity Theorem. Find the ratios of the lengths of the corresponding sides. = = = = 4 SR SP 16 4 SP + PR SP = = = = 4 ST SQ 20 5 SQ + QT SQ Because S is the included angle in both triangles, use the SAS Similarity Theorem to conclude that  RST ~  PSQ. The side lengths SR and ST are proportional to the corresponding side lengths of  PSQ PQ S RT

U SING S IMILAR T RIANGLES IN R EAL L IFE S CALE D RAWING As you move the tracing pin of a pantograph along a figure, the pencil attached to the far end draws an enlargement. Using a Pantograph P R T S Q

U SING S IMILAR T RIANGLES IN R EAL L IFE Using a Pantograph As the pantograph expands and contracts, the three brads and the tracing pin always form the vertices of a parallelogram. P R T S Q

U SING S IMILAR T RIANGLES IN R EAL L IFE Using a Pantograph The ratio of PR to PT is always equal to the ratio of PQ to PS. Also, the suction cup, the tracing pin, and the pencil remain collinear. P R T S Q

You know that. Because P  P, you can apply the SAS Similarity Theorem to conclude that  PRQ ~  PTS. = PQ PS PR PT Using a Pantograph S R Q T P How can you show that  PRQ ~  PTS ? S OLUTION

Because the triangles are similar, you can set up a proportion to find the length of the cat in the enlarged drawing. Using a Pantograph 10 " 2.4 " 10 " S R Q T P = RQ TS PR PT Write proportion. = TS = 4.8 Substitute. Solve for TS. So, the length of the cat in the enlarged drawing is 4.8 inches. In the diagram, PR is 10 inches and RT is 10 inches. The length of the cat, RQ, in the original print is 2.4 inches. Find the length TS in the enlargement.

Finding Distance Indirectly Similar triangles can be used to find distances that are difficult to measure directly. R OCK C LIMBING You are at an indoor climbing wall. To estimate the height of the wall, you place a mirror on the floor 85 feet from the base of the wall. Then you walk backward until you can see the top of the wall centered in the mirror. You are 6.5 feet from the mirror and your eyes are 5 feet above the ground. 85 ft6.5 ft 5 ft A B C E D Use similar triangles to estimate the height of the wall. Not drawn to scale

Finding Distance Indirectly 85 ft6.5 ft 5 ft A B C E D Use similar triangles to estimate the height of the wall. S OLUTION Using the fact that  ABC and  EDC are right triangles, you can apply the AA Similarity Postulate to conclude that these two triangles are similar. Due to the reflective property of mirrors, you can reason that ACB  ECD.

85 ft6.5 ft 5 ft A B C E D  DE Finding Distance Indirectly Use similar triangles to estimate the height of the wall. S OLUTION = EC AC DE BA Ratios of lengths of corresponding sides are equal. Substitute. Multiply each side by 5 and simplify. DE 5 = So, the height of the wall is about 65 feet.