Cfand aeb afc oto prove abc isosceles abe and acf

Triangle

A

BC

“CPCT” means corresponding parts of congruent triangles.

B C E F If in ∆sABC and DEF, AB = DE, AC = DF and ∠BAC = ∠EDF
Then, ABC≅ ∆DEF . It is called SAS congruence rule i.e. side-angle-side]

To Prove:ABC≅ ∆DEF Fig.1

Proof:Case – (i)
Let AB = DE [Assumed]

But it is given that ∠ACB = ∠DFE
∴∠ACB = ∠PCB which is not possible.

i.e. it is only possible if P coincides with A
⇒ BA = ED
Hence, ABC≅ ∆DEF [by SAS rule]
Case – (iii)
If AB < DE, we can choose a point M on DE such that ME = AB and repeating the arguments as given in case (ii), we can conclude that AB = DE and, hence, ABC≅ ∆DEF Now, two triangles are congruent if any two pairs of angles and one pair of corresponding sides are equal. We may call it as Angle-Angle-Side (AAS).

In ∆ABC altitudes BE = CF
To prove: (i) ABE≅ ∆ACF
(ii) AB = AC ; i.e. ABC is an isosceles triangle.

Solution:
(i) In ∆ABE and ∆ ACF, we have
BE = CF [Given]
∠AEB = ∠AFC = 90o [ BE and CF are altitudes: Given] and ∠A = ∠A [Common]

C E

A D

BC EF If ABC and DEF are two right triangles in which
∠B = ∠E = 90o, AC = DF and AB = DE

AB
Given:
ABC is a right angled triangle in which ∠A = 90o and AB = AC.

To find: ∠B and ∠C
Solution:
In right angled triangle ABC, we have
AB = AC [Given]
∴∠B = ∠C
[Angles opposite to equal sides of a triangle are equal] ....(i) Now, ∠A + ∠B + ∠C = 180o [Angles sum property of a ∆] ⇒ 90o + ∠B + ∠B = 180o [From (i)
∠B = ∠C]
⇒ 2∠B = 180o – 90o⇒ 2∠B = 90o

Hence, ∠B = ∠C = 45o

Construction: AD bisector of ∠A is drawn at BC i.e. ∠BAD = ∠CAD Proof: In ∆BAD and ∆CAD, we have
AB = AC [Given]
∠BAD = ∠CAD [By construction]
and AD = AD [Common]

B C Let ABC be a triangle in which AC > AB and AC > BC.

The side opposite to the largest angle is the longest.

LEARN IT
Theorem: The sum of any two sides of a triangle is greater than the third side.

Solution:
For ∆ABC and ∆ADC, we have AB = AD
∠BAC = ∠DAC
AC = AC
Therefore, by SAS criteria, we have ∆ABC ≅∆ADC
Also, we have corresponding parts of the two triangle equal and it gives BC = DC.

Solution:
Let us consider ∆PAB and ∆PAC (as shown in figure 7.35) Here, we have PB = PC (Given)
∠PBA = ∠PCA (Each = 90º)
PA = PA (Common side)
Then by RHS congruence criteria, We have
∆PAB ≅∆PAC.

 An exterior angle of a triangle is Now, ∠ADC > ∠ABD ... (2)  greater than an opposite interior angle From (1) and (2)
∠ACD > ∠ABD
i.e., ∠ACD > ∠ABC
i.e., in ∆ABC,
∠C > ∠B
⇒ AB > AC

⇒ AB > AD ….(1)
Similarly, in right angled ∆ABC and ∆CFA, we have BC > BE ….(2)
….(3) and CA > CF
Adding (1), (2), (3)
AB + BC + CA > AD + BE + CF
i.e., Perimeter of ∆ABC > (AD + BE + CF).

Illustration-10:
AD is an altitude of an isosceles triangle ABC in which AB = AC. Show that

Solution:
AD is an altitude from vertex A on BC
∴∠ADB = ∠ADC = 90o
(i) Now, in right ∆sADB and ADC ,
we have
∠ADB = ∠ADC = 90o [Given]
and AD = AD [Common]
∴ ∆ADB≅ ∆ADC

[By RHS congruence rule]
∴ BD = CD [By CPCT]
Hence, AD bisects BC. Proved.

KEY POINTS

4. The sum and difference of two angles of a triangle are 128° and 22° respectively. Find all the angles of the triangle.

A

(D) 1/2(∠R).

8.	In figure, triangle ABC is right-angled at B. Given that AB = 9 cm, AC = 15 cm, calculate the length of BC	A		C
		B

9. The triangle formed by joining the mid-points of the sides of an isosceles triangles is

(A) scalene triangle (C) right triangle

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