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Question 3,4 & 5


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Answer 1

3. Triangle ABC with AB = AC and D and E are points BC such that BE = CD. Then show that AD = AE

Here, image is not given so two cases maybe there,

SOLUTION:

• CASE 1

[See the attached image (1)]

AB = AC (given)    -----(i)

So, ΔABC is an isosceles triangle.

∴ ∠ABC = ACB     -------(ii)

Now,

BE = CD (given) --------(iii)

Taking eq.(i), (ii), and (iii), we get

ΔABE ≅ ΔACD

By SAS - Side Angle Side.

Hence, by CPCT, AD = AE (proved)

- - -

• CASE 2

[See image 2, there will be the only small change in name but the solution will remain the same.]

AB = AC (given)    -----(1)

So, ΔABC is an isosceles triangle.

∴ ∠ABC = ACB     -------(2)

Now,

BE = CD (given) --------(3)

Taking eq.(1), (2), and (3), we get

ΔABE ≅ ΔACD

By SAS - Side Angle Side.

Hence, by CPCT, AD = AE (proved)

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4. In the given figure, find the value of ∠DBC+∠ACE+∠BAF, if ABC is a triangle.

SOLUTION:

[See image 3(attached)]

To make to easy to understand let the name

∠DBC = ∠y

∠ACE = ∠z

∠BAF = ∠x

And

interior angle ∠ABC = ∠c

interior angle ∠BCA = ∠a

interior angle ∠CAB = ∠b

Now,

We know that,

Sum of all three interior angles of a triangle = 180° [Angle sum property]

So,

∠a + ∠b + ∠c = 180°

We get

∠z + ∠a = 180°,

∠y + ∠c = 180°,

∠x + ∠b = 180°

(Linear pairs)

So,

(∠z + ∠a = 180°) + (∠y + ∠c = 180°) + (∠x + ∠b = 180°)

⇒ ∠z + ∠a + ∠y + ∠c + ∠x + ∠b = 180° + 180° + 180°

⇒ ∠z + ∠a + ∠y + ∠c + ∠x + ∠b = 540°

⇒ ∠a + ∠b + ∠c + ∠z + ∠x + ∠y = 540°

[as ∠a + ∠b + ∠c  = 180° (interior angles of the triangle ABC) ]

⇒ 180° + ∠z + ∠x + ∠y = 540°

⇒ ∠z + ∠x + ∠y = 540° - 180°

⇒ ∠z + ∠x + ∠y = 360°

Hence,

∠DBC+∠ACE+∠BAF = 360°

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5. In a triangle(ABC), BD is perpendicular (bisector) to AC and angle B = 90°. Then Show that BD = 1/2 AC.

SOLUTION:

BD is perpendicular to AC.

Also, BD bisects ∠B.

So ∆ABD & ∆BDC will be right angle Isosceles triangle.

So, ∠DAB = ∠ABD = ∠DBC = ∠DCB = 45°

Thus, AD = BD = CD

So, AD = BD = CD

Therefore. BD = AD

BD + CD = AD + CD ( Adding CD on both sides)

BD + BD = AC (Since, BD = CD)

2BD = AC

BD = 1/2 AC

Answer 2

Attachment of the solution 3,4&5

Answer 3 :-

Given :-

In ΔABC,

D & E are points on BC, such that BE = CD

AD = AE

To Prove :-

AD = AE

Solution :-

AD = AE

In, ΔABE & Δ ACD

• AB = AC

• ∠B = ∠C

• BE = CD

By, SAS ΔABC is congruent to ΔACD.

By CPCT

AD = AE

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Answer 4 :-

In fig. find the value of ∠DBC+∠ACE+∠BAF in ΔABC.

Let ,

• ∠DBC = x

• ∠ACE = y

• ∠BAF = z

Then,

• ∠ ABC = 180-x

• ∠BCA = 180-y

• ∠CAB = 180 -z

We, know sum of angles is 180°.

∠ ABC + ∠BCA + ∠CAB = 180°

➡ 180-x + 180-y + 180 -z = 180

➡ 540 - (x+y+z) = 180

➡ x+y+z = 540-180

➡ x+y+z = 360°

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Answer 5 :-

In ΔADB = ΔABC,

∠ D = ∠ B

∠ A = ∠ A. (common)

AA similarity,

$\frac{BD}{BC}$ = $\frac{AB}{AC}$

BD = $\frac{AB}{AC}$ × BC

$\frac{AD}{AB}$ = $\frac{AB}{BC}$

ADB is congruent to BDC.

Hence, AD = BC

Ar. Δ ABC = 2. Ar. ΔABD

=> $\frac{1}{2}$ × BC × AB = 2 × $\frac{1}{2}$ × BD × AD

=> $\frac{1}{2}$ × BC × AB = BD × AD

= > BD = $\frac{1}{2}$ AC

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