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Standard VII

Mathematics

ASA Criteria for Congruency

Line l is the...

Question

Line l is the bisector of an angle $∠ A a n d B$ is any point on l. BP and BQ are perpendiculars from B to the arms of $∠ A$ (see the given figure). Show that:
$Δ A P B ≅ Δ A Q B$

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Solution

$Δ A P B a n d Δ A Q B ∠ A P B = ∠ A Q B (E a c h 90^{\circ}) ∠ P A B = ∠ Q A B (l i s a n g l e b i s e c t o r o f ∠ A) A B = A B (C o m m o n) ∴ Δ A P B ≅ Δ A Q B (B y A A S C o n g r u e n c e R u l e)$

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Similar questions

Q. Line l is the bisector of an angle

∠ A a n d B

is any point on l. BP and BQ are perpendiculars from B to the arms of

∠ A

(see the given figure). Show that:

Δ A P B ≅ Δ A Q B

Question 5 (i)
Line l is the bisector of an angle $∠ A a n d B$ is any point on l. BP and BQ are perpendiculars from B to the arms of $∠ A$ (see the given figure). Show that:
$Δ A P B ≅ Δ A Q B$

Q. Line

l

is the bisector of an angle

∠ A

and

∠ B

is any point on

l

B P

and

B Q

are perpendicular from

B

to the arms of

∠ A

(see figure). Show that
(i)

Δ A P B = Δ A Q B

(ii)

B P = B Q

B

is equidistant from the arms of

∠ A

Q. In the given figure, Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A. Show that :
(i) ΔAPB ≅ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠A.

Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that:

(i) ΔAPB ≅ ΔAQB

(ii) BP = BQ or B is equidistant from the arms of ∠A.

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Line l is the bisector of an angle ∠A and B is any point on l. BP and BQ are perpendiculars from B to the arms of ∠A (see the given figure). Show that: ΔAPB≅ΔAQB

ΔAPB and ΔAQB∠APB=∠AQB(Each 90∘)∠PAB=∠QAB(l is angle bisector of ∠A)AB=AB(Common)∴ΔAPB≅ΔAQB(By AAS Congruence Rule)

Line l is the bisector of an angle $∠ A a n d B$ is any point on l. BP and BQ are perpendiculars from B to the arms of $∠ A$ (see the given figure). Show that:
$Δ A P B ≅ Δ A Q B$

$Δ A P B a n d Δ A Q B ∠ A P B = ∠ A Q B (E a c h 90^{\circ}) ∠ P A B = ∠ Q A B (l i s a n g l e b i s e c t o r o f ∠ A) A B = A B (C o m m o n) ∴ Δ A P B ≅ Δ A Q B (B y A A S C o n g r u e n c e R u l e)$