AP State Syllabus AP Board 9th Class Maths Solutions Chapter 7 Triangles Ex 7.1 Textbook Questions and Answers.

## AP State Syllabus 9th Class Maths Solutions 7th Lesson Triangles Exercise 7.1

Question 1.

In quadrilateral ACBD, AC = AD and AB bisects ∠A. Show that ΔABC ≅ ΔABD What can you say about BC and BD ?

Solution:

Given that AC = AD

∠BAC = ∠BAD (∵ AB bisects∠A)

Now in ΔABC and ΔABD

AC = AD (∵ given)

∠BAC = ∠BAD (Y given)

AB = AB (common side)

∴ ΔABC ≅ ΔABD

(∵ SAS congruence rule)

Question 2.

ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA, prove that i) ΔABD ≅ΔBAC ii) BD = AC

iii) ∠ABD = ∠BAC.

Solution :

i) Given that AD = BC and

∠DAB = ∠CBA

Now in ΔABD and ΔBAC

AB = AB (∵ Common side)

AD = BC (∵ given)

∠DAB = ∠CBA (∵ given)

∴ ΔABD ≅ ΔBAC

(∵ SAS congruence)

ii) From (i) AC = BD (∵ CPCT)

iii) ∠ABD = ∠BAC [ ∵ CPCT from (i)]

Question 3.

AD and BC are equal and perpendi-culars to a line segment AB. Show that CD bisects AB.

Solution:

Given that AD = BC; AD ⊥ AB; BC ⊥ AB

In ΔBOC and ΔAOD

∠BOC = ∠AOD (∵ vertically opposite angles)

∴ ΔOBC = ΔOAD (∵ right angle)

BC = AD

ΔOBC ≅ ΔOAD (∵ AAS congruence)

∴ OB = OA (∵ CPCT)

∴ ‘O’ bisects AB

Also OD = OC

∴ ‘O’ bisects CD

⇒ AB bisects CD

Question 4.

l and m are two parallel lines inter-sected by another pair of parallel lines p and q. Show that ΔABC ≅ ΔCDA.

Solution:

Given that l // m; p // q.

In ΔABC and ΔCDA

∠BAC = ∠DCA (∵ alternate interior angles)

∠ACB = ∠CAD

AC = AC

∴ ΔABC ≅ ΔCDA (∵ ASA congruence)

Question 5.

In the figure given below AC = AE; AB = AD and ∠BAD = ∠EAC. Show that BC = DE.

Solution:

Given that AC = AE, AB = AD and

∠BAD = ∠EAC

In ΔABC and ΔADE

AB = AD

AC = AE

∠BAD = ∠EAC

∴ ΔABC ≅ ΔADE (∵ SAS congruence)

⇒ BC = DE (CPCT)

Question 6.

In right triangle ABC, right angle is at ‘C’ M is the mid-point of hypotenuse AB. C is joined to M and produced to a point D such that DM = CM. Point D is joined to point B (see fig.). Show that

i) ΔAMC = ΔBMD

ii) ∠DBC is a right angle

iii) ΔDBC = ΔACB

iv) CM = \(\frac{1}{2}\) AB

Solution:

Given that ∠C = 90°

M is mid point of AB;

DM = CM (i.e., M is mid point of DC)

i) In ΔAMC and ΔBMD

AM = BM (∵ M is mid point of AB)

CM = DM ( ∵ M is mid point of CD)

∠AMC = ∠BMD ( ∵ Vertically opposite angles)

∴ ΔAMC ≅ ΔBMD

(∵ SAS congruence)

ii) ∠MDB = ∠MCA

(CPCT of ΔAMC and ΔBMD)

But these are alternate interior angles for the lines DB and AC and DC as transversal.

∴DB || AC

As AC ⊥ BC; DB is also perpendicular to BC.

∴ ∠DBC is a right angle.

iii) In ΔDBC and ΔACB

DB = AC (CPCT of ΔBMD and ΔAMC)

∠DBC = ∠ACB = 90°(already proved)

BC = BC (Common side)

∴ ΔDBC ≅ ΔACB (SAS congruence rule)

iv) DC = AB (CPCT of ΔDBC and ΔACB)

\(\frac { 1 }{ 2 }\) DC = \(\frac { 1 }{ 2 }\) AB (Dividing both sides by 2)

CM = \(\frac { 1 }{ 2 }\)AB

Question 7.

In the given figure ΔBCD is a square and ΔAPB is an equilateral triangle.

Prove that ΔAPD ≅ ΔBPC.

[Hint: In ΔAPD and ΔBPC; \(\overline{\mathbf{A D}}=\overline{\mathbf{B C}}\), \(\overline{\mathbf{AP}}=\overline{\mathbf{BP}}\) and ∠PAD = ∠PBC = 90° – 60° = 30°]

Solution:

Given that □ABCD is a square.

ΔAPB is an equilateral triangle.

Now in ΔAPD and ΔBPC

AP = BP ( ∵ sides of an equilateral triangle)

AD = BC (∵ sides of a square)

∠PAD = ∠PBC [ ∵ 90° – 60°]

∴ ΔAPD ≅ ΔBPC (by SAS congruence)

Question 8.

In the figure given below ΔABC is isosceles as \(\overline{\mathbf{A B}}=\overline{\mathbf{A C}} ; \overline{\mathbf{B A}}\) and \(\overline{\mathbf{CA}}\) are produced to Q and P such that \(\overline{\mathbf{A Q}}=\overline{\mathbf{AP}}\). Show that \(\overline{\mathbf{PB}}=\overline{\mathbf{QC}}\) .

(Hint: Compare ΔAPB and ΔACQ)

Solution:

Given that ΔABC is isosceles and

AP = AQ

Now in ΔAPB and ΔAQC

AP = AQ (given)

AB = AC (given)

∠PAB = ∠QAC (∵ Vertically opposite angles)

∴ ΔAPB ≅ ΔAQC (SAS congruence)

∴ \(\overline{\mathbf{PB}}=\overline{\mathbf{QC}}\) (CPCT of ΔAPB and ΔAQC)

Question 9.

In the figure given below AABC, D is the midpoint of BC. DE ⊥ AB, DF ⊥ AC and DE = DF. Show that ΔBED ≅ AΔCFD.

Solution:

Given that D is the mid point of BC of ΔABC.

DF ⊥ AC; DE = DF

DE ⊥ AB

In ΔBED and ΔCFD

∠BED = ∠CFD (given as 90°)

BD = CD (∵D is mid point of BC)

ED = FD (given)

∴ ΔBED ≅ ΔCFD (RHS congruence)

Question 10.

If the bisector of an angle of a triangle also bisects the opposite side, prove that the triangle is isosceles.

Solution:

Let ΔABC be a triangle.

The bisector of ∠A bisects BC

To prove: ΔABC is isosceles

(i.e., AB = AC)

We know that bisector of vertical angle divides the base of the triangle in the ratio of other two sides.

∴ \(\frac{\mathrm{AB}}{\mathrm{AC}}=\frac{\mathrm{BD}}{\mathrm{BC}}\)

Thus \(\frac{\mathrm{AB}}{\mathrm{AC}}\) = 1( ∵ given)

⇒ AB = AC

Hence the Triangle is isosceless.

Question 11.

In the given figure ΔABC is a right triangle and right angled at B such that ∠BCA = 2 ∠BAC. Show that the hypotenuse AC = 2BC.

[Hint : Produce CB to a point D that BC = BD]

Solution:

Given that ∠B = 90°; ∠BCA = 2∠BAC

To prove : AC = 2BC

Produce CB to a point D such that

BC = BD

Now in ΔABC and ΔABD

AB = AB (common)

BC = BD (construction)

∠ABC =∠ABD (∵ each 90°)

∴ ΔABC ≅ ΔABD

Thus AC = AD and ∠BAC = ∠BAD = 30° [CPCT]

[ ∵ If ∠BAC = x then

∠BCA = 2x

x + 2x = 90°

3x = 90°

⇒ x = 30°

∴ ∠ACB = 60°]

Now in ΔACD,

∠ACD = ∠ADC = ∠CAD = 60°

∴∠ACD is equilateral ⇒ AC = CD = AD

⇒ AC = 2BC (∵ C is mid point)