Solved Examples

SOLVED EXAMPLES 1. In figure 5-19, AE bisects ∠CAD and ∠B = ∠C . Prove that AE parallel to BC . Sol: Since is the bisector of exterior angle of Triangle ABC

Therefore ∠CAD = ∠B + ∠C It is given that ∠B = ∠C

Therefore ∠CAD = ∠B + ∠B ∠CAD = 2∠B

… (1)

Since AE bisects ∠CAD ∠CAD = 2∠DAE

… (2)

From (1) and (2), we find that

2∠B = 2∠DAE

(Equals of equals are equal)

∠B = ∠DAE

Therefore They are corresponding angles

Hence AE parallel to BC 2. In a right triangle, one acute angle is double the other. Prove that the hypotenuse is double the smallest side. Sol:

Given: In Triangle ABC , ∠B = 90o , ∠ACB = 2∠BAC To prove: AC=2BC Construction: Produce CB to D such that BD=BC. Join AD> Proof: In Triangle ABC and Triangle ABD we have BC=BD

(By Construction)

∠ABC = ∠ABD = 90o Since, AB is the common side to both the triangles,

Triangle ABC ≅ Triangle ABD

(SAS Congruence)

Hence, AC=AD and ∠BAC = ∠BAD triangles are congruent) Let ∠BAC = x o

(Corresponding

parts

of

congruent

Then ∠BAD = x o

( Since ∠BAC = ∠BAD proved above)

Thus ∠ACB = 2∠BAC

(given)

∠ACB = 2 x o ∠DAC = ∠BAD + ∠BAC = x o + x o = 2 x o = ∠ACB Therefore AD = DC (sides opposite to equal angles are equal) or AC=DC

( Since AD = AC )

or AC=2BC Note: This is a special case of a right triangle when the angles are 30o , 60o ,90o 3. Prove that medians of an equilateral triangle are equal. Sol:

Given: Triangle ABC is an equilateral triangle and AD, BE and CF are its medians To prove: AD=BE=CF Proof: In Triangle ABD and Triangle CBF , we have AB=AC

(Given)

BD=BF

(Halves of equals are equal)

∠B is common

Therefore Triangle ABD ≅ Triangle CBF (SAS Congruence) or AD=CF

(Corresponding parts of congruent triangle are congruent)

Similarly, proceeding the same way, it can be show that AD=BE Hence AD=BE=CF 4. If the altitude from the two vertices of a triangle to the opposite sides are equal, prove that the triangle is isosceles. Sol:

Given: Triangle ABC in which BE ⊥ AC and CF ⊥ AB such that BE=CF To prove: AB=AC Proof: IN Triangle ABE and Triangle ACF , we have

∠AEB = ∠AFC = 90o

(Given)

∠A = ∠A

(Common to both the triangles)

BE=CF

(Given)

Therefore Triangle ABE ≅ Triangle ACF congruence)

(By

SAA

criterion

of

Thus, AB=AC (cpctc) Hence, Triangle ABC is isosceles (Note: Wherever cpctc is written, it means corresponding parts of congruent triangles are congruent) 5. In figure 5-23, AB=AC D is the point in the interior of Triangle ABC such that ∠DBC = ∠DCB . Prove that AD bisects ∠BAC of Triangle ABC Sol:

Given: Triangle ABC in which AB=AC ∠DBC = ∠DCB

To prove: ∠BAD = ∠CAD Proof: In Triangle BDC , we have ∠DBC = ∠DCB

(Given)

Thus, DC=DB

… (1)

(Sides opposite to equal angles of a triangle are equal) Now, in Triangle ABD and Triangle ACD , we have AB=AC

(Given)

BD=CD

(From 1)

and AD=AD

(Common side to both the triangles)

So, by SSS congruence criterion, we have

Triangle ABD ≅ Triangle ACD ∠BAD = ∠CAD

Hence, AD is the bisector of ∠BAC 6. In figure 5-24, AD=BC and BC=CA. Prove that ∠ADB = ∠BCA and ∠DAB = ∠CBA Sol:

Given: Quadrilateral ABCD in which AD=BC and BD=CA To prove: 1. ∠ADB = ∠BCA 2. ∠DAB = ∠BAC Proof: In Triangle ABD and Triangle ABC , we have AD=BC

(Given)

BD=CA

(Given)

and AB=AB

(Common side to both the triangles)

By SSS criterion of congruence, we have Triangle ABD ≅ Triangle BAC Hence, ∠DAB = ∠CBA Therefore ∠ADB = ∠BCA

(cpctc) ( Since ∠DAB = ∠CBA )

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