# Ex 7.1 NCERT Solutions- Triangles Class 9 Notes | EduRev

## Class 9 : Ex 7.1 NCERT Solutions- Triangles Class 9 Notes | EduRev

The document Ex 7.1 NCERT Solutions- Triangles Class 9 Notes | EduRev is a part of the Class 9 Course Mathematics (Maths) Class 9.
All you need of Class 9 at this link: Class 9

Question 1. In quadrilateral ACBD, AC = AD and AB bisects ∠A (see figure). Show that ΔABC ≌ ΔABD. What can you say about BC and BD?
Solution:
and AB being the bisector of ∠A.
Now, in ΔABC and ΔABD, AB = AB               [Common]
∠CAB = ∠DAB               [∵ AB bisects ∠ CAD]
∴ Using SAS criteria, we have ΔABC ≌ ΔABD.
∵ Corresponding parts of congruent triangles (c.p.c.t) are equal.
∴ BC = BD.

Question 2. ABCD is a quadrilateral in which AD = BC and ∠DAB = ∠CBA (see Figure). Prove that
(i)
ΔABD ≌ ΔBAC (ii) BD = AC (iii) ∠ ABD = ∠ BAC.
Solution:
(i) In quadrilateral ABCD, we have AD = BC and ∠DAB = ∠CBA.
In ΔABD and ΔBAC, AB = BA              [Common]
∠DAB = ∠CBA              [Given]
∴ Using SAS criteria, we have ΔABD ≌ ΔBAC

(ii) ∵ ΔABD ≌ ΔBAC
∴ Their corresponding parts are equal.
⇒ BD = AC

(iii) Since ΔABD ≌ ΔBAC
∴ Their corresponding parts are equal.
⇒ ∠ABD = ∠BAC.

Question 3. AD and BC are equal perpendiculars to a line segment AB (see figure). Show that CD bisects AB.
Solution:
We have ∠ABC = 90° and ∠BAD = 90°
Also AB and CD intersect at O.
∴ Vertically opposite angles are equal.
Now, in ΔOBC and ΔOAD, we have ∠ABC = ∠BAD       [each = 90°]
∠BOC = ∠AOD              [vertically opposite angles]
∴ Using ASA criteria, we have ΔOBC ≌ ΔOAD
⇒ OB = OA              [c.p.c.t.]
i.e. O is the mid-point of AB Thus, CD bisects AB.

Question 4. l and m are two parallel lines intersected by another pair of parallel lines p and q (see figure). Show that ΔABC ≌ ΔCDA. Solution: ∵ ℓ || m and AC is a transversal,
∴ ∠BAC = ∠DCA              [Alternate interior angles]
Also p || q and AC is a transversal,
∴ ∠BCA = ∠DAC              [Alternate interior angles]
Now, in ΔABC and ΔCDA,
∠BAC = ∠DCA              [Proved]
∠BCA = ∠DAC              [Proved]
CA = AC               [Common]
∴ Using ASA criteria, we have ΔABC ≌ ΔCDA

Question 5. 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 Figure). Show that:
(i)
ΔAPB ≌ ΔAQB
(ii) BP = BQ or B is equidistant from the arms of ∠ A. Solution: We have, ℓ as the bisector of QAP.
∴ ∠QAB = ∠PAB
∠Q= ∠P              [each = 90°]
⇒ Third ∠ABQ = Third ∠ABP (i)
Now, in ΔAPB and ΔAQB, we have
AB = AB              [common]
∠ABP = ∠ABQ               [proved]
∠PAB = ∠QAB              [proved]
∴ Using SAS criteria, we have ΔAPB ≌ ΔAQB

(ii) Since ΔAPB ≌ ΔAQB
∴ Their corresponding angles are equal.
⇒ BP = BQ
i.e. [Perpendicular distance of B from AP] = [Perpendicular distance of B from AQ]
Thus, the point B is equidistant from the arms of ∠A.

Question 6. In the figure, AC = AE, AB = AD and ∠BAD = ∠EAC. Show that BC = DE. Solution: We have ∠BAD = ∠EAC
Adding ∠DAC on both sides, we have
∠BAD + ∠DAC = ∠EAC + ∠DAC
⇒ ∠BAC = ∠DAE
Now, in ΔABC and ΔADE, we have
∠BAC = ∠DAE              [Proved]
AC = AE              [Given]
∴ ΔABC ≌ ΔADE              [Using SAS criteria]
Since ΔABC ≌ ΔADE, therefore, their corresponding parts are equal.
⇒ BC = DE.

Question 7. AB is a line segment and P is its mid-point. D and E are points on the same side of AB such that ∠BAD = ∠ABE and ∠EPA = ∠DPB (see figure). Show that (i) ΔDAP ≌ ΔEBP (ii) AD = BE Solution: We have, P is the mid-point of AB.
∴ AP = BP
∠EPA = ∠DPB               [Given]
Adding ∠EPD on both sides, we get
∠EPA + ∠EPD = ∠DPB + ∠EPD
⇒ APD = ∠BPE

(i) Now, in ΔDAP ≌ ΔEBP, we have
AP = BP.              [Proved]
∠DPA = ∠EPB              [Proved]
∴ Using ASA criteria, we have
ΔDAP ≌ΔEBP

(ii) Since, ΔDAP ≌ΔEBP
∴ Their corresponding parts are equal. ⇒ AD = BE.

Question 8. In right triangle ABC, right angled 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 figure). Show that (i) ΔAMC ≌ ΔBMD (ii) ∠ DBC is a right angle. (iii) ΔDBC ≌ ΔACB (iv) CM = (1/2)AB
Solution:
∵ M is the mid-point of AB.
∴ BM = AM              [Given]

(i) In ΔAMC and ΔBMD, we have CM = DM               [Given]
AM = BM              [Proved]
∠AMC = ∠BMD               [Vertically opposite angles]
∴ ΔAMC ≌ ΔBMD               (SAS criteria)

(ii) ∵ ΔAMC ≌ ΔBMD
∴ Their corresponding parts are equal.
⇒ ∠MAC = ∠MBD But they form a pair of alternate interior angles.
∴ AC || DB
Now, BC is a transversal which intersecting parallel lines AC and DB,
∴ ∠BCA + ∠DBC = 180°
But ∠BCA = 90               [∵ ΔABC is right angled at C]
∴ 90° + ∠DBC = 180°
or ∠DBC = 180° ∠ 90° = 90°
Thus, ∠DBC = 90°

(iii) Again, ΔAMC ≌ ΔBMD               [Proved]
∴ AC = BD              [c.p.c.t]
Now, in ΔDBC and ΔACB, we have
∠DBC = ∠ACB               [Each = 90°]
BD = CA              [Proved]
BC = CB              [Common]
∴ Using SAS criteria, we have ΔDBC ≌ ΔACB

(iv) ∵ ΔDBC ≌ ΔACB
∴ Their corresponding parts are equal.
⇒ DC = AB
But DM = CM               [Given]
∴ CM = (1/2) DC = (1/2) AB
⇒ CM = (1/2) AB

Offer running on EduRev: Apply code STAYHOME200 to get INR 200 off on our premium plan EduRev Infinity!

## Mathematics (Maths) Class 9

190 videos|233 docs|82 tests

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

,

;