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**Solve the last exercise Chapter 9 Math 7 Connect**

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### Solve lesson 9.36 page 84 Math 7 textbook Connecting knowledge volume 2 – KNTT

Let ABC be a triangle \(\widehat{BAC}\) an obtuse angle. Take point D between A and B, take point E between A and C (H.9.51). Prove DE < BC.

Detailed instructions for solving problem 9.36

**Solution method**

DC>DE (relation between angle and opposite side in triangle DEC). (first)

Consider triangle ADC => \(\widehat{BDC}\) is an obtuse angle (2)

From (1) and (2) infer: BC>DE

**Detailed explanation**

Since \(\widehat{BAC}\) is an obtuse angle, \(\widehat{ADE}\) , \(\widehat{AED}\) are acute angles

=> \(\widehat{DEC}\) is an obtuse angle.

=>DC>DE (relation between angle and opposite side in triangle DEC). (first)

Consider triangle ADC with:

\(\widehat{DAC}\) is an obtuse angle so \(\widehat{ADC}\) , \(\widehat{ACD}\) are acute angles

=> \(\widehat{BDC}\) is an obtuse angle.

=>BC>DC (relationship between angle and opposite side in triangle BDC) (2)

– From (1) and (2) infer: BC>DE

–>

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### Solve problem 9.37 page 84 Math textbook 7 Connecting knowledge volume 2 – KNTT

Let ABC be a triangle (AB > AC). On the line containing side BC, take point D and point E such that B lies between BD and C, C lies between B and E, BD = BA, CE = CA (H.9.52)

a) Compare \(\widehat{ADE}\) and \(\widehat{AED}\)

b) Compare the line segments AD and AE

Detailed instructions for solving problem 9.37

**Solution method**

a) AB > AC => \(\widehat{ABC}\) < \(\widehat{ACB}\), find the measure \(\widehat{ACE}\), \(\widehat{ABD}\) then compare

Triangle ABD is isosceles at B ( BD= BA) => \(\widehat{ABD}\) = 180°- 2\(\widehat{ADB}\)

Triangle ACE is isosceles at C ( CE= CA) => \(\widehat{ACE}\) = 180°- 2\(\widehat{AEC}\)

Find the measure of two angles \(\widehat{ADB}\), \(\widehat{AEC}\) and compare

b) Consider triangle ADE

**Detailed explanation**

a) AB > AC => \(\widehat{ABC}\) < \(\widehat{ACB}\) (relation between angle and opposite side in triangle ABC)

\(\widehat{ABD}\) + \(\widehat{ABC}\) = 180° => \(\widehat{ABC}\) = 180°- \(\widehat{ABD}\)

\(\widehat{ACE}\) + \(\widehat{ACB}\) = 180° => \(\widehat{ACB}\) = 180°- \(\widehat{ACE}\)

=> 180°- \(\widehat{ABD}\) < 180°- \(\widehat{ACE}\)

=> \(\widehat{ACE}\) < \(\widehat{ABD}\)

Triangle ABD is isosceles at B ( BD= BA) => \(\widehat{ABD}\) = 180°- 2\(\widehat{ADB}\)

Triangle ACE is isosceles at C ( CE= CA) => \(\widehat{ACE}\) = 180°- 2\(\widehat{AEC}\)

=> 180°- 2\(\widehat{ADB}\) > 180°- 2\(\widehat{AEC}\)

=> \(\widehat{ADB}\) < \(\widehat{AEC}\)

b) Considering triangle ADE, we have: \(\widehat{ADB}\) < \(\widehat{AEC}\)

=> AD > AE

–>

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### Solve problem 9.38 page 84 Math 7 textbook Connecting knowledge volume 2 – KNTT

Let AI and AM be the altitude and median, respectively, from the vertex A of triangle ABC. Prove that

a) AI < \(\frac{1}{2}\) (AB + AC)

b) AM < \(\frac{1}{2}\) (AB + AC)

Detailed instructions for solving problem 9.38

**Solution method**

a) AI is the altitude from A to the line segment BC

=> AI < AB and AI < AC

Adding the two sides together we have => AI <\(\frac{1}{2}\) (AB + AC)

b) Take D such that M is the midpoint of AD

Considering ABM and DCM, prove that ABM = DCM

**Detailed explanation**

a) AI is the altitude from A to the line segment BC=> AI is the distance from A to BC => AI is the shortest

=> AI < AB and AI < AC

Adding the two sides together we have: 2 AI < AB + AC

=> AI <\(\frac{1}{2}\) (AB + AC)

b) Take D such that M is the midpoint of AD

Consider ABM and DCM have

AM = DM (M is mid point of AD)

BM=CM (M is mid point of BC)

\(\widehat{AMB}\) = \(\widehat{CMD}\) (2 opposite angles)

=> ABM = DCM

=> AB = CD

Considering ADC we have: AD < AC + CD

=> 2AM < AC + AB

=> AM < \(\frac{1}{2}\) (AB + AC)

–>

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### Solve problem 9.39 page 84 Math 7 textbook Connecting knowledge volume 2 – KNTT

Let ABC be a triangle with bisectors AD and D lying on BC such that BD = 2 DC. On the line AC, take point E such that C is the midpoint of AE (H.9.53). Prove that triangle ABE is isosceles at A

Hint D is the centroid of the triangle ABE, which has the bisector AD and the median.

Detailed instructions for solving problem 9.39

**Solution method**

D is the centroid of triangle ABE, which has the bisector AD and is the median.

**Detailed explanation**

C is the midpoint of AE => BC is the median of triangle ABE (1)

D belongs to BC, BD= 2DC

=> BC= BD + DC = 2DC + DC = 3DC => DC = \(\frac{1}{3}\) BC (2)

From (1) and (2)=> D is the centroid of triangle ABE

=> AD is the median line with BE

where AD is the bisector of \(\widehat{BAC}\) or \(\widehat{BAE}\) of triangle ABE

=> Triangle ABE is isosceles at A

–>

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