Exercise 7.1

1.In quadrilateral ACBD, AC = AD and AB bisects  

$$∠A$$ Show that  $$ΔABC≅  ΔABD$$ . What can you say about BC and BD?

Sol.Given that AC=AD

$$\angle BAC =\angle BAD$$   ( $$\because$$ AB   bisects $$\angle a$$ )

Now in

$$\Delta ABC$$ and $$\Delta ABD$$

AC=AD (

$$\because$$ Given)

$$\angle$$ BAC= $$\angle BAD$$ ( $$\because$$ given)

AB=AB (common side)

$$\therefore \Delta ABC\cong \Delta ABD$$

(

$$\because$$ SAS congruence rule)

2.ABCD is a quadrilateral in which AD = BC and

$$∠DAB = ∠CBA$$ Prove that (i) $$ΔABD≅  ΔBAC$$ (ii) BD = AC (iii) $$∠ABD = ∠BAC$$ .

Sol.i)Given that AD=BC and

$$\angle DAB=\angle CBA$$

Now in

$$\Delta ABD$$ and $$\Delta BAC$$

AB=AB (

$$\because$$ common side) 

AD=BC (

$$\because$$ given)

$$\angle DAB=\angle CBA$$ ( $$\because$$ given)

$$\therefore \Delta$$ ABD $$\cong \Delta BAC$$

(

$$\because$$   SAS congruence) 

ii)From i)AC=BD (

$$\because$$ CPCT)

iii)

$$\angle ABD=\angle$$ BAC [ $$\because$$ CPCT from (1)]

3.AD and BC are equal and perpendiculars to a line segment AB.  Show that CD bisects AB.

Sol.Given that

$$AD=BS$$ ; $$AD\perp AB;BC\perp AB in \Delta BOC$$ and $$\Delta AOD$$

$$\angle BOC=\angle AOD$$ ( $$\because$$ vertically opposite angles)

$$\angle OBC=\angle OAD$$ ( $$\because$$ right angle)

BC=AD

(

$$\therefore  \Delta OBC\cong \Delta OAD$$ ) ( $$\because AAS$$ )

$$\therefore$$ 'O'  bisects AB

Also OD=OC

$$\therefore$$ 'O' bisects CD

$$\Rightarrow$$ AB bisects CD

4.l and m are two parallel lines intersected by another pair of parallel lines p and q . Show that 

$$ΔABC≅  ΔCDA$$

Sol.Given that l

$$\parallel$$ m; p $$\parallel$$ q.

In

$$\Delta ABC and \Delta CDA$$

$$\angle BAC=\angle DCA$$ ( $$\because$$ alternate interior angles)

$$\angle ACB=\angle CAD$$

AC=AC

$$\therefore \Delta ABC\cong \Delta CDA$$ ( $$\because$$ ASA congruence)

5.In the adjacent figure,  AC = AE, AB = AD and

$$\angle BAD = \angle EAC$$ . Show that BC = DE.

Sol.Given that AC=AE,AB=AD and

$$\angle BAD=\angle EAC$$

$$In \Delta ABC$$ and $$\Delta ADE$$

AB=AE

AC=AE

$$\angle BAD=\angle EAC$$

$$\therefore \Delta ABC\cong \Delta ADE$$ ( $$\because$$ SAS congruence)

$$\Rightarrow BC = DE(CPCT)$$

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 figure).  Show that 

i)

$$ΔAMC≅  ΔBMD$$

(ii)

$$∠ DBC$$ is a right angle 

(iii)

$$ΔDBC ≅  ΔACB$$

(iv)

$$CM=\frac{1}{2}AB$$

Sol.Given that

$$\angle c=90^{\circ}$$

M is mid point of AB;

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

i)In

$$\Delta AMC$$ and $$\Delta BMD$$

AM=BM (

$$\because$$ M is mid point of AB)

CM=DM (

$$\because$$   M is mid point of CD)

$$\angle AMC=\angle BMD$$ ( $$\because$$ vertically opposite angles)

$$\therefore \Delta AMC\cong \Delta BMD$$ ( $$\because$$ SAS congruence)

ii)

$$\angle MDB=\angle MCA$$ (CPCT of $$\Delta AMC$$ and $$\Delta BMD$$ )

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

$$\therefore DB\parallel AC$$

As

$$AC \perp BC ; DB$$ is also perpendicular to BC.

$$\therefore \angle DBC$$ is a right angle.

iii)in

$$\Delta DBC$$ and $$\Delta ACB$$

DB=AC (CPCT of

$$\Delta BMD$$ and $$\Delta AMC$$ )

$$\angle DBC=\angle ACB 90^{\circ}$$ (already proved)

BC=BC(common side)

$$\therefore \Delta DBC\cong \Delta ACB$$ (SAS congruence rule)

iv)DC=AB (CPCT of

$$\Delta DBC$$ and $$\Delta ACB$$ ) $$\frac{1}{2}DC = \frac{1}{2}AB$$ (Dividing both sides by 2) 

CM=

$$\frac{1}{2} AB$$

7.In the adjacent figure ABCD is a square and

$$ΔAPB$$ is an equilateral triangle.  Prove that $$ΔAPD≅  ΔBPC$$ . [Hint : In $$ΔAPD$$ and $$ΔBPC  \overline{AD}= \overline{BC}$$ , $$\overline{AP}$$ = $$\overline{BP}$$ and $$∠PAD$$ = $$∠ PBC = 90^{\circ} - 60^{\circ}$$ = $$30^{\circ}]$$

Sol.Given that ABCD is a square.

$$\Delta APB$$ is an equilateral triangle.

Now in

$$\Delta APD$$ and $$\Delta BPC$$

AP=BP (

$$\because$$ sides of an equilateral triangle)

AB=BC (

$$\because$$ sides of a square)

$$\angle PAD=\angle PBC$$   [ $$\because 90^{\circ}-60^{\circ}$$ ]

$$\therefore \Delta APD\cong \Delta BPC$$ (by SAS congruence)

8.In the adjacent figure   

$$ΔABC$$ is isosceles as $$\overline{AB}=\overline{AC}$$ ; $$\overline{BA}$$ and $$\overline{CA}$$ are produced to Q and P such that $$\overline{AQ} =\overline{AP}$$ .Show that $$\overline{PB}=\overline{QC}$$ .

(Hint:compare

$$\Delta APB$$ and $$\Delta ACQ$$ )

Sol.Given that

$$\Delta ABC$$ is isosceles and AP=AQ

Now in

$$\Delta APB$$ and $$\Delta AQC$$

AP=AQ (given)

AB=AC (given)

$$\angle PAB=\angle QAC$$ ( $$\because$$ vertically opposite angles)

$$\therefore \Delta APB\cong \Delta AQC$$ (SAS congruence)

$$\therefore \overline{PB}=\overline{QC}$$ (CPCT of $$\Delta APB$$ and $$\Delta AQC$$ )

9.In the figure given below

$$\Delta ABC$$ , D is the midpoint of BC. $$DE\perp AB,DF\perp AC$$ and DE=DF. show that $$\Delta BED\cong \Delta CFD$$ .

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

$$\Delta ABC$$ .

$$DF\perp$$ AC;DE=DF

$$DE\perp$$ AB

In

$$\Delta BED$$ and $$\Delta CFD$$

$$\angle BED=\angle CFD$$ (given as $$90^{\circ}$$ )

BD=CD (

$$\because$$ D is mid point of BC)

ED=FD (given)

$$\therefore \Delta BED\cong \Delta$$ CFD(RHS congruence)

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

Sol.Let

$$\Delta ABC$$ be a triangle.

the bisector of

$$\angle$$ A bisects BC

To prove:

$$\Delta ABC$$ 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.

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

Thus

$$\frac{AB}{AC}$$ =1 ( $$\because$$ given)

$$\Rightarrow$$ AB=AC

Hence the triangle is isosceles.

11.In the given figure ABC is a right triangle and right angled at B such that

$$\angle ABC = 2 \angle BAC$$ . Show that hypotenuse AC = 2BC.

(Hint : Produce CB to a point D that BC = BD)

Sol.Given that

$$\angle B=90^{\circ};\angle BCA=2\angle BAC$$

To prove:AC=2BC

Produce CB to a point D such that

BC=BD

Now in

$$\Delta ABC$$ and $$\Delta ABD$$

AB=AB (common)

BC=BD (construction)

$$\angle ABC=\angle ABD$$ ( $$\because Each 90^{\circ}$$ )

$$\therefore \Delta ABC\cong ABD$$

Thus AC=AD and

$$\angle BAC=\angle BAD=30^{\circ}[CPCT]$$

$$[\because If \angle BAC=X then \angle BCA=2X$$

$$X+2x=90^{\circ}$$

$$3x=90^{\circ}$$

$$\Rightarrow X=30^{\circ}$$

$$\therefore \angle ACB=60^{\circ}]$$

Now in

$$\Delta ACD$$

$$\angle ACD=\angle ADC=\angle$$ CAD= $$60^{\circ}$$

$$\therefore \angle ACD$$ is equilateral $$\Rightarrow AC=CD=AD$$

$$\Rightarrow AC=2BC$$ ( $$\because$$ C is mid point)

 

 

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