Question

CAN SOMEBODY ANS ME THOSE QUESTIONS. IN ENGLISH . ️️wrong ans will be reported and the correct one will be the branliast and...... Try to give the and correct

Answer 1

Question 1:

Given:

⇒ A quadrilateral ABCD.

⇒ AD = BC

⇒ ∠DAB = ∠CBA

To Prove:

⇒ (i) ΔABD ≅ ΔBAC

⇒ (ii) BD = AC

⇒ (iii) ∠ABD = ∠BAC

Solution:

(i) In ΔABD and ΔBAC,

AB = AB (Common side)

∠DAB = ∠CBA (Given)

AD = BC (Given)

∴ By using SAS Congruency rule,

ΔABD ≅ ΔBAC

(ii) We know that ΔABD ≅ ΔBAC

∴ By using CPCT (Corresponding Parts of Congruent Triangles) we can say that:

BD = AC

(iii) We know that ΔABD ≅ ΔBAC

∴ By using CPCT (Corresponding Parts of Congruent Triangles) we can say that:

∠ABD = ∠BAC

Hence proved.

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Question 2:

Given:

⇒ AD = BC

⇒ AD ⊥ BA

⇒ CB ⊥ BA

⇒ ∠DAO = ∠CBO = 90°

To Prove:

⇒ CD bisects BA

In simple words, we have to prove that BO = OA

Solution:

In ΔAOD and ΔBOC,

∠DAO = ∠CBO = 90° (Given)

∠DOA = ∠COB (Vertically opposite angles are equal)

AD = BC (Given)

∴ By using AAS Congruency rule we can say that,

ΔAOD ≅ ΔBOC

∴ By using CPCT (Corresponding Parts of Congruent Triangles) we can say that:

BO = OA

I.e, CA bisects BA.

Hence Proved.

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Question 3:

Given:

⇒ l ║m

⇒ p║q

To Prove:

⇒ ΔABC ≅ ΔCDA

Solution:

ATQ, l║m, and considering AC as the transversal we can say that:

∠CAD = ∠ACB (Alternate angles are equal)

ATQ, p║q, and considering AC as the transversal we can say that:

∠BAC = ∠DCA (Alternate angles are equal)

Now, In ΔABC and ΔCDA,

∠CAD = ∠ACB (Proved)

∠BAC = ∠DCA (Proved)

AC = AC (Common side)

∴ By using AAS Congruency rule we can say that,

ΔABC ≅ ΔCDA

Hence proved.

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Question 4:

Given:

⇒ $\ell$ bisects ∠A

⇒ B is a point on $\ell$.

⇒ BP ⊥ AP

⇒ BQ ⊥ AQ

To Prove:

(i) ΔAPB ≅ ΔAQB

(ii) BQ = BQ

Solution:

$\ell$ bisects ∠A

⇒ ∠QAB = ∠PAB ⇢ (1)

BP ⊥ AP

⇒ ∠BPA = 90° ⇢ (2)

BQ ⊥ AQ

⇒ ∠BQA = 90° ⇢ (3)

From (2) and (3) we can say that:

⇒ ∠BPA = ∠BQA = 90° ⇢ (4)

[Things that are equal to the same thing are equal to one another]

Now in ΔAPB and ΔAQB,

∠QAB = ∠PAB (Proved in 1)

∠BPA = ∠BQA (Proved in 4)

AB = AB (Common side)

∴ By using AAS Congruency rule we can say that,

ΔAPB ≅ ΔAQB

∴ By using CPCT (Corresponding Parts of Congruent Triangles) we can say that:

BP = BQ

Hence proved.