Mid-point and Its Converse (Including Intercept Theorem) — ICSE Class 9 Maths Important Questions

Answer: (b) $4.5$4.5 cm.

By the mid-point theorem $DE=\dfrac12 BC=\dfrac12\times9=4.5$DE=12 BC=12×9=4.5 cm.

Answer: (a) The mid-point of $PR$PR.

By the converse of the mid-point theorem, a line through the mid-point of one side parallel to another side bisects the third side, so $T$T is the mid-point of $PR$PR.

Answer: (c) Parallelogram.

Each mid-segment is parallel to and half a diagonal of the quadrilateral. Both pairs of opposite mid-segments become parallel and equal, so the figure is always a parallelogram.

Answer: (b) $3.2$3.2 cm.

By the intercept theorem, if parallel lines make equal intercepts on one transversal they make equal intercepts on any transversal. So the second intercept equals the first, $3.2$3.2 cm.

Answer: (a) By the mid-point theorem $EF=\tfrac12 BC$EF=12 BC, $FD=\tfrac12 CA$FD=12 CA, $DE=\tfrac12 AB$DE=12 AB. Adding, perimeter of $\triangle DEF=\tfrac12$DEF=12(perimeter of $\triangle ABC$ABC). R correctly explains A.

By the mid-point theorem, $DE\parallel BC$DE BC and $DE=\dfrac12 BC$DE=12 BC.

Therefore $BC=2\times DE=2\times3.5=7$BC=2× DE=2×3.5=7 cm.

Relation: $DE$DE is parallel to $BC$BC and equal to half of $BC$BC.

$F$F is the mid-point of $AB$AB and $E$E is the mid-point of $AC$AC.

By the mid-point theorem, $EF\parallel BC$EF BC and $EF=\dfrac12 BC$EF=12 BC.

Therefore $EF=\dfrac12\times10=5$EF=12×10=5 cm, and $EF$EF is parallel to $BC$BC.

By the mid-point theorem:

$FD\parallel AC$FD AC and $FD=\dfrac12 AC=AE=EC$FD=12 AC=AE=EC (since $E$E is mid-point of $AC$AC).

$DE\parallel AB$DE AB and $DE=\dfrac12 AB=AF=FB$DE=12 AB=AF=FB (since $F$F is mid-point of $AB$AB).

$EF\parallel BC$EF BC and $EF=\dfrac12 BC=BD=DC$EF=12 BC=BD=DC.

In $\triangle FBD$FBD and $\triangle DEF$DEF:
$FB=DE$FB=DE, $BD=EF$BD=EF, $FD=FD$FD=FD (common).

By SSS, $\triangle FBD\cong\triangle DEF$FBD DEF.

In quadrilateral $AFDE$AFDE: $AF\parallel ED$AF ED (as $DE\parallel AB$DE AB) and $AE\parallel FD$AE FD (as $FD\parallel AC$FD AC). Both pairs of opposite sides are parallel, so $AFDE$AFDE is a parallelogram.

Join $A$A to $C$C, meeting $MN$MN at $P$P.

In $\triangle ADC$ADC, $M$M is the mid-point of $AD$AD and $MP\parallel DC$MP DC (as we will show $MN\parallel DC\parallel AB$MN DC AB). By the converse of the mid-point theorem $P$P is the mid-point of $AC$AC, and by the mid-point theorem

$MP=\dfrac12 DC.$MP=12 DC.

In $\triangle ABC$ABC, $P$P is the mid-point of $AC$AC and $N$N is the mid-point of $BC$BC, so by the mid-point theorem

$PN=\dfrac12 AB,\quad PN\parallel AB.$PN=12 AB, PN AB.

Since $MP\parallel DC\parallel AB$MP DC AB and $PN\parallel AB$PN AB, the points $M,P,N$M,P,N are collinear and $MN\parallel AB$MN AB.

Adding, $MN=MP+PN=\dfrac12 DC+\dfrac12 AB=\dfrac12(AB+DC)$MN=MP+PN=12 DC+12 AB=12(AB+DC). Hence proved.

Since $ABCD$ABCD is a parallelogram, $AB\parallel DC$AB DC and $AB=DC$AB=DC.

As $E$E and $F$F are mid-points, $AE=\dfrac12 AB$AE=12 AB and $FC=\dfrac12 DC$FC=12 DC. Hence $AE=FC$AE=FC and $AE\parallel FC$AE FC, so $AECF$AECF is a parallelogram, giving $DE\parallel BF$DE BF... more precisely $EB\parallel DF$EB DF and $EB=DF$EB=DF, so $DEBF$DEBF is a parallelogram and $DE\parallel BF$DE BF.

Let $DE$DE and $BF$BF cut diagonal $AC$AC at $P$P and $Q$Q.

In $\triangle ABQ$ABQ (wait, use $\triangle$s directly): In $\triangle ABP$ABP... apply the converse of the mid-point theorem. In $\triangle ABQ$ABQ where $Q$Q lies on $AC$AC: $E$E is the mid-point of $AB$AB and $EP\parallel BQ$EP BQ (since $DE\parallel BF$DE BF). By the converse of the mid-point theorem, $P$P is the mid-point of $AQ$AQ, so $AP=PQ$AP=PQ.

Similarly, in $\triangle DPC$DPC (with the parallels), $F$F is the mid-point of $DC$DC and $FQ\parallel DP$FQ DP, so $Q$Q is the mid-point of $PC$PC, giving $PQ=QC$PQ=QC.

Therefore $AP=PQ=QC$AP=PQ=QC, i.e. $DE$DE and $BF$BF trisect the diagonal $AC$AC. Hence proved.

Given: $\triangle ABC$ABC in which $D$D and $E$E are the mid-points of $AB$AB and $AC$AC.

To prove: $DE\parallel BC$DE BC and $DE=\dfrac12 BC$DE=12 BC.

Construction: Produce $DE$DE to $F$F so that $DE=EF$DE=EF, and join $CF$CF.

Proof: In $\triangle AED$AED and $\triangle CEF$CEF:

$AE=CE$AE=CE ($E$E is mid-point of $AC$AC)

$\angle AED=\angle CEF$AED= CEF (vertically opposite angles)

$DE=EF$DE=EF (by construction)

By SAS, $\triangle AED\cong\triangle CEF$AED CEF.

Hence $AD=CF$AD=CF and $\angle ADE=\angle CFE$ADE= CFE (c.p.c.t.).

The equal alternate angles $\angle ADE=\angle CFE$ADE= CFE imply $AD\parallel CF$AD CF, i.e. $DB\parallel CF$DB CF.

Now $AD=CF$AD=CF and $AD=DB$AD=DB (as $D$D is mid-point of $AB$AB), so $DB=CF$DB=CF.

Thus $DB$DB and $CF$CF are equal and parallel, so $DBCF$DBCF is a parallelogram. Therefore $DF\parallel BC$DF BC and $DF=BC$DF=BC.

Since $D,E,F$D,E,F are collinear, $DE\parallel BC$DE BC.

Also $DF=DE+EF=2DE$DF=DE+EF=2DE, so $2DE=BC$2DE=BC, giving $DE=\dfrac12 BC$DE=12 BC. Hence proved.

Join diagonal $AC$AC.

In $\triangle ABC$ABC, $P$P and $Q$Q are mid-points of $AB$AB and $BC$BC. By the mid-point theorem:

$PQ\parallel AC \text{ and } PQ=\dfrac12 AC \quad\text{...(1)}$PQ AC and PQ=12 AC ...(1)

In $\triangle ADC$ADC, $S$S and $R$R are mid-points of $AD$AD and $CD$CD. By the mid-point theorem:

$SR\parallel AC \text{ and } SR=\dfrac12 AC \quad\text{...(2)}$SR AC and SR=12 AC ...(2)

From (1) and (2): $PQ\parallel SR$PQ SR and $PQ=SR$PQ=SR.

Since one pair of opposite sides is equal and parallel, $PQRS$PQRS is a parallelogram.

Special case: Join diagonal $BD$BD. In $\triangle ABD$ABD, $PS=\dfrac12 BD$PS=12 BD; in $\triangle CBD$CBD, $QR=\dfrac12 BD$QR=12 BD. So $PS=QR=\dfrac12 BD$PS=QR=12 BD.

If $AC=BD$AC=BD, then $PQ=\dfrac12 AC=\dfrac12 BD=PS$PQ=12 AC=12 BD=PS. Thus two adjacent sides of the parallelogram are equal, so all four sides are equal.

Therefore, when the diagonals are equal, $PQRS$PQRS is a rhombus.

(i) By the mid-point theorem each mid-segment is half the opposite side:

$DE\parallel CA$DE CA, $DE=\dfrac12 CA=\dfrac12\times100=50$DE=12 CA=12×100=50 m.

$EF\parallel AB$EF AB, $EF=\dfrac12 AB=\dfrac12\times60=30$EF=12 AB=12×60=30 m.

$FD\parallel BC$FD BC, $FD=\dfrac12 BC=\dfrac12\times80=40$FD=12 BC=12×80=40 m.

(ii) $EF$EF is parallel to $AB$AB and equal to half of $AB$AB.

(iii) Perimeter of $\triangle DEF=50+30+40=120$DEF=50+30+40=120 m.

(iv) Perimeter of $\triangle ABC=60+80+100=240$ABC=60+80+100=240 m. Fraction $=\dfrac{120}{240}=\dfrac12$=120/240=12, i.e. one half.