## Math 7 Chapter 2 Lesson 8: Congruent cases of right triangles

## 1. Summary of theory

### 1.1. Known congruent cases of two right triangles

– If the two right-angled sides of one right triangle are equal to the two sides of the other triangle, then the two right triangles are congruent (cgc).

If a right angled side and an adjacent acute angle of one right triangle are equal to one side of this right triangle and an adjacent acute angle of another right triangle, then those two right triangles equal (gcg)

– If the hypotenuse and an acute angle of one right triangle are equal to the hypotenuse and an acute angle of the other right triangle, then the two right triangles are congruent (gcg).

### 1.2. The case of equal hypotenuse and right angle

If the hypotenuse and one right angled side of one right triangle are equal to the hypotenuse and one right angled side of another right triangle, then the two triangles are congruent.

## 2. Illustrated exercise

**Question 1: **Given triangle ABC is isosceles at A, draw AD perpendicular to BC. Prove that AD is the bisector of angle A.

**Solution guide**

Consider two right triangles ADB and ADC with common side AD

AB = AC (gt)

So \(\Delta ADB = \Delta ADC\) (hype – right angle side)

Derive \(\widehat {BAD} = \widehat {CAD}\) (corresponding angle)

So AD is the bisector of angle A.

**Verse 2: **Given an isosceles triangle ABC at A, draw \(BH \bot AC,CK \bot AB.\) Let I be the intersection of BH and CK. Prove that AI is the bisector of angle A.

**Solution guide**

Considering two triangles AHB and AKC, we have: AB=AC (gt)

\({B_1} = \widehat {{C_1}}\) (together \(\frac{1}{2}\widehat B = \frac{1}{2}\widehat C\))

So \(\Delta AHB = \Delta AKC\) (hype, acute angle) infer AH = AK (corresponding side)

Considering two right triangles AHI and AKI, we have:

Common edge AI

AH= AK (CM above)

So \(\Delta AHI = \Delta AKI\) (hype, right angle side)

So \(\widehat {{A_1}} = \widehat {{A_2}}\)

So AI is the bisector of angle A.

**Question 3:** Let \(\Delta ABC\) square at A. On the side BC take a point D such that BD = BA. The line perpendicular to BC at D intersects AC at E.

a. Compare the lengths of AE and DE

b. The bisector of the exterior angle at vertex C intersects the line BE at K. Calculate \(\widehat {BAK}.\)

**Solution guide**

a. Join BE consider \(\Delta ABE\) and \(\Delta DBE,\) to have:

\(\begin{array}{l}BAE = BDE = {90^0}\\BA = BD\,\,(gt)\end{array}\)

BC common edge

So \(\Delta ABE = \Delta DBE\) (case of hypotenuse, right angle side)

Deduce AE = DE

b. Connect AK

Since \(\Delta ABE = \Delta DBE\) we have \(\widehat {ABE} = \widehat {DBE}\) or BK is the bisector of angle B. Draw \(KM \bot BC,\, \,KN \bot AB,\,KH \bot AC.\)

Two right triangles KHC and KMC have common hypotenuse KC, two equal acute angles \(\widehat {KCH} = \widehat {KCM}\) (CK is the bisector of \(\widehat {HCM}\)) so \(\Delta KHC = \Delta KMC\)

So KH = KM

Similarly \(\Delta KNB = \Delta KMB\) (hypotenuse, acute angle)

So KM = KN

Infer that KH = KN (same as KM)

Consider two right triangles KAH and KAN with:

KA common edge

KH = KN

So \(\Delta KAH = \Delta KAN\) (hype, right angle side)

So \(\widehat {{A_1}} = \widehat {{A_2}} = \frac{1}{2}\widehat {HAN} = {45^0}\)

Therefore

\(\begin{array}{l}\widehat {BAK} = \widehat {BAC} + \widehat {{A_2}}\\ = {90^0} + {45^0}\end{array}\)

So \(\widehat {BAK} = {135^0}\)

## 3. Practice

### 3.1. Essay exercises

**Question 1: **Given \(\Delta ABC\) square at A (AB < AC), M is a point on side AC. Draw MH perpendicular to BC \((H \in BC)\). Know MH = HB. Prove that AH is the bisector of angle A.

**Verse 2:** Let ABC be an isosceles triangle with base BC. From B draw a line perpendicular to AB and from C draw a line perpendicular to AC. These two lines intersect at M. Prove that:

a. \(\Delta ABM = \Delta ACM\)

b. AM is the perpendicular bisector of BC.

**Question 3:** Let \(\Delta ABC\) square at A. In the outer region of the triangle draw right triangles ABD, ACF (AB = BD; AC = CF).

a. Prove that D, A, F are collinear.

b. From D and F lower the perpendiculars DD’, FF’ to the line BC. Prove: DD’ + FF’ = BC.

### 3.2. Multiple choice exercises

**Question 1: **Given triangle ABC and triangle NPM with BC = PM, \(\widehat B = \widehat P = {90^0}\). What additional conditions are needed for triangle ABC and triangle NPM to be congruent in the case of hypotenuse – right angle side?

A. BA = PM

B. BA = PN

C. CA = MN

D. \(\widehat A = \widehat N\)

**Verse 2: **Let triangle ABC and triangle MNP have \(\widehat A = \widehat M = {90^0},\widehat C = \widehat P\). What additional condition is needed for triangle ABC and triangle MNP to be congruent in the case of right angle side – adjacent acute angle?

A. AC = MP

B. AB = MN

C. BC = NP

D. AC = MN

**Question 3: **Let triangle ABC and triangle DEF have \(\widehat B = \widehat E = {90^0},AC = DF,\widehat A = \widehat F\). Which of the following statements is correct?

A. \(\Delta ABC = \Delta F{\rm{ED}}\,\,\)

B. \(\Delta ABC = \Delta F{\rm{DE}}\,\,\)

C. \(\Delta BAC = \Delta F{\rm{ED}}\,\,\)

D. \(\Delta ABC = \Delta D{\rm{EF}}\,\,\)

**Question 4: **Given triangle ABC and triangle WHEN there are: \(\widehat A = \widehat K = {90^0},AB = KH,BC = HI\). Which of the following statements is correct?

A. \(\Delta ABC = \Delta WHEN\)

B. \(\Delta ABC = \Delta HIK\)

C. \(\Delta BAC = \Delta WHEN\)

D. \(\Delta ACB = \Delta WHEN\)

**Question 5: **Given triangle ABC and triangle DEF with AB = DE, \(\widehat B = \widehat E,\widehat A = \widehat D = {90^0}\). Know AC = 9cm. The length of DF is:

A. 10cm

B. 5cm

C. 9cm

D. 7cm

## 4. Conclusion

Through this lesson, you should know the following:

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