9th Maths 7.1

Exercise 7.1

Q1. In quadrilateral $ABCD$ , $AC=AD$ and $AB$ bisects $angleA$ (see given figure). Show that $triangleABCcongtriangleABD$ . What can you say about $BC$ and $BD$ ?

Sol. :

$triangleABC$ and

$triangleABD$

$AC=AD$ [Given]

$angleCAB=angleBAD$ [

$AB$ is bisect

$angleA$ ]

$AB=AB$ [Common]

$triangleABCcongtriangleABD$ [By

$SAS$ rule]

$BC=BD$ [

$CPCT$ ]

Proved.

Q2. $ABCD$ is a quadrilateral in which $AD=BC$ and $angleDAB=angleCBA$ (see given figure). Prove that

i) $triangleABCcongtriangleBAC$

ii) $BD=AC$

iii) $angleABD=angleBAC$ .

Sol. ;

i) In

$triangleABD$ and

$triangleBAC$

$AD=BC$ [Given]

$angleDAB=angleCBA$ [Given]

$AB=AB$ [Common]

$triangleABDcongtriangleBAC$ [By

$SAS$ rule]

ii)

$triangle ABCcongtriangleBAC$ [Proved in (i)]

$BD=AC$ [CPCT]

Proved.

iii)

$triangleABDcongtriangleBAC$ [Proved in (i)]

$angleABD=angleBAC$ [CPCT]

Proved.

Q3. $AD$ and $BC$ are equal perpendicular to a line segment $AB$ (see given figure). Show that $CD$ bisects $AB$ .

Sol. ;

$triangleAOD$ and

$triangleBOC$

$BC=AD$ [Given]

$angleDAO=angleCBO=90^circ$ [Given

$BC⊥AB$ ,

$AD⊥AB$ ]

$angleDOA=angleCOB$ [Vertically opposite angles]

$triangleAODcongtriangleBOC$ [By

$S$

$A$

$A$ rule]

$AO=OB$ [CPCT]

So that

$CD$ bisect

$AB$ .

Proved

Q4. $l$ and $m$ are two parallel lines $p$ and $q$ (see given figure). Show that $triangleABCcongtriangleCDA$ .

Sol. :

$triangleABC$ and

$triangleADC$

$l∥m$ and

$AC$ is transversal.

$angleACB=angleDAC$ [Alternate angles]

$AC=AC$ [Common Side]

$p∥q$ and

$AC$ is transversal.

$angleBAC=angleACD$ [Alternate angles]

$triangleABCcongtriangleADC$ [By

$ASA$ rule]

Proved.

Q5. Line $l$ is the bisector of an angle $angleA$ and $B$ is any point on $l$ . $BP$ and $BQ$ are perpendiculars from $B$ to the arms of $angleA$ (see given figure). Show that:

i) $trangleAPBcongtriangleAQB$

ii) $BP=BQ$ or $B$ is equidistant from the arms of $angleA$ .

Sol. :

i) In

$triangleAPB$ and

$triangleAQB$

$angleBAP=angleBAQ$ [Line

$l$ bisect

$angleA$ ]

$angleAPB=angleAQB$ [

$BP⊥AP$ and

$BQ⊥AQ$ ]

$AB=AB$ [Common]

$triangleAPBcongtriangleAOB$ [By

$S$

$A$

$A$ rule]

ii)

$triangleAPBcongtriangleAQB$ [Proved in (i)]

$BP=BQ$ [CPCT]

Proved.

Q6. In the given figure $AC=AE, AB=AD$ and $angleBAD=angleEAC$ . Show that $BC=DE$ .

Sol. :

$triangleABC$ and

$triangleADE$

$AB=AD$ [Given]

$AC=AE$ [Given]

$angleBAD=angleEAC$ [Given]

We add

$angleDAC$ both sides]

$angleBAD+angleDAC=angleEAC+angleDAC$

$angleBAC=angleDAE$

$triangleABCcongtriangleADE$ [By

$SAS$ rule]

$BC=DE$ [CPCT]

Proved.

Q7. $AB$ is a line segment and $p$ is its mid-point. $D$ and $E$ are points on the same side of $AB$ such that $angleBAD=angleABE$ and $angleEPA=angleDPB$ (see given figure). Show that

i) $triangleDAPcongtriangleEBP$

ii) $AD=BE$

Sol. :

i) In

$triangleDAP$ and

$triangleEBP$

$AP=PB$ [

$P$ is mid point of

$AB$ ]

$angleAPE=angleBPD$ [Given ]

We add

$angleEPD$ both sides

$angleAPE+angleEPD=angleBPD+angleEPD$

$angleAPD=angleBPE$

$angleDAP=angleEBP$ [Given]

$triangleDAPcongtriangleEBP$ [By

$ASA$ rule]

ii)

$triangleDPAcongtriangleEBP$ [Proved in (i)]

$AD=BE$ [CPCT]

Proved.

Q8. 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 given figure). Show that :

i) $triangleAMCcongtriangleBMD$

ii) $angleDBC$ is a right angle`

iii) $triangleDBCcongtriangleACB$

iv) $CM=frac(1)(2)AB$ .

Sol. :

i) In

$triangleAMC$ and

$triangleBMD$

$AM=BM$ [Given]

$CM=DM$ [Given]

$angleCMA=angleBMD$ [Vertically opposite angles]

$triangleAMCcongtriangleBMD$ [By

$SAS$ rule]

Proved.

ii)

$triangleAMCcongtriangle$ [Proved in (i)]

$angleACM=angleBDM$ [CPCT]

above angles are Alternate angles,

so that

$DB∥AC$

$angleDBC+angleACB=180^circ$ [interior angles of same side]

$angleDBC+90^circ=180^circ$ [

$angleACB=90circ$ ]

$angleDBC=180^circ-90^circ$

So that

$angleDCB$ is a right angle.

Proved.

iii)

$triangleAMCcongtriangleBMD$ [Proved in (i)]

$AC=BD$ [CPCT]

$triangleDBC$ and

$triangleACB$

$BD=AC$ [Proved]

$angleDBC=angleACB=90^circ$ [Proved in (ii)]

$BC=BC$ [Common]

$triangleDBCcongtriangleACB$ [By

$SAS$ rule]

Proved.

iv)

$triangleDBCcongtriangleACB$ [Proved in (iii)]

$DC=AB$ [CPCT]

$CM+MD=AB$

$CM+CM=AB$ [Given

$CM=MD$ ]

$2CM=AB$

$CM=1/2AB$

Proved.

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